OP Malhotra Class 12 Maths Solutions Chapter 16 Definite Integrals Exercise 16 (B)

Question 1. Evaluate the following definite integrals:
(i) \( \int_1^2 \frac{3 x}{9 x^2-1} d x \)
(ii) \( \int_0^1 \frac{x^5}{1+x^6} d x \)
(iii) \( \int_0^1 \frac{x}{1+x^4} d x \)
(iv) \( \int_0^1 \frac{e^{2 x}}{1+e^{4 x}} d x \)
(v) \( \int_1^2 3 x \sqrt{5-x^2} d x \)
(vi) \( \int_1^2 \frac{e^{-\frac{1}{x}}}{x^2} d x \)
(vii) \( \int_0^a x \sqrt{a^2-x^2} d x \)
(viii) \( \int_0^{(\pi / 2)^{1 / 3}} x^2 \sin x^3 d x \)
Answer:
(i) Let \( I = \int_1^2 \frac{3 x d x}{9 x^2-1} \)
To solve this, we use substitution. Let \( t = 9x^2 - 1 \).
Then, find the derivative of \( t \) with respect to \( x \): \( dt = 18x dx \).
This means \( x dx = \frac{1}{18} dt \).
Next, change the limits of integration. When \( x = 1 \), \( t = 9(1)^2 - 1 = 8 \).
When \( x = 2 \), \( t = 9(2)^2 - 1 = 9(4) - 1 = 36 - 1 = 35 \).
Now, rewrite the integral in terms of \( t \):
\( I = \int_8^{35} \frac{3}{t} \left( \frac{1}{18} dt \right) \)
\( I = \int_8^{35} \frac{1}{6t} dt \)
\( I = \frac{1}{6} [\log |t|]_8^{35} \)
Apply the limits of integration:
\( I = \frac{1}{6} (\log 35 - \log 8) \)
Using logarithm properties \( \log a - \log b = \log (a/b) \):
\( I = \frac{1}{6} \log \left( \frac{35}{8} \right) \)
(ii) Let \( I = \int_0^1 \frac{x^5 d x}{1+x^6} \)
For this integral, we use substitution. Let \( t = 1+x^6 \).
Differentiate \( t \) with respect to \( x \): \( dt = 6x^5 dx \).
This gives \( x^5 dx = \frac{1}{6} dt \).
Next, change the limits. When \( x = 0 \), \( t = 1+0^6 = 1 \).
When \( x = 1 \), \( t = 1+1^6 = 2 \).
Substitute these into the integral:
\( I = \int_1^2 \frac{1}{t} \left( \frac{1}{6} dt \right) \)
\( I = \frac{1}{6} \int_1^2 \frac{1}{t} dt \)
\( I = \frac{1}{6} [\log |t|]_1^2 \)
Apply the limits:
\( I = \frac{1}{6} (\log 2 - \log 1) \)
Since \( \log 1 = 0 \):
\( I = \frac{1}{6} \log 2 \)
(iii) Let \( I = \int_0^1 \frac{x d x}{1+x^4} \)
We use substitution for this integral. Let \( t = x^2 \).
Differentiate \( t \) with respect to \( x \): \( dt = 2x dx \).
So, \( x dx = \frac{1}{2} dt \).
Change the limits of integration. When \( x = 0 \), \( t = 0^2 = 0 \).
When \( x = 1 \), \( t = 1^2 = 1 \).
Substitute these into the integral:
\( I = \int_0^1 \frac{1}{1+t^2} \left( \frac{1}{2} dt \right) \)
\( I = \frac{1}{2} \int_0^1 \frac{1}{1+t^2} dt \)
Recognize this as a standard integral form: \( \int \frac{1}{1+t^2} dt = \tan^{-1} t \).
\( I = \frac{1}{2} [\tan^{-1} t]_0^1 \)
Apply the limits:
\( I = \frac{1}{2} (\tan^{-1} 1 - \tan^{-1} 0) \)
We know \( \tan^{-1} 1 = \frac{\pi}{4} \) and \( \tan^{-1} 0 = 0 \).
\( I = \frac{1}{2} \left( \frac{\pi}{4} - 0 \right) = \frac{\pi}{8} \)
(iv) Let \( I = \int_0^1 \frac{e^{2 x}}{1+e^{4 x}} d x \)
Use substitution here. Let \( t = e^{2x} \).
Differentiate \( t \) with respect to \( x \): \( dt = 2e^{2x} dx \).
So, \( e^{2x} dx = \frac{1}{2} dt \). Note that \( e^{4x} = (e^{2x})^2 = t^2 \).
Change the limits. When \( x = 0 \), \( t = e^{2(0)} = e^0 = 1 \).
When \( x = 1 \), \( t = e^{2(1)} = e^2 \).
Substitute these into the integral:
\( I = \int_1^{e^2} \frac{1}{1+t^2} \left( \frac{1}{2} dt \right) \)
\( I = \frac{1}{2} \int_1^{e^2} \frac{1}{1+t^2} dt \)
\( I = \frac{1}{2} [\tan^{-1} t]_1^{e^2} \)
Apply the limits:
\( I = \frac{1}{2} (\tan^{-1} (e^2) - \tan^{-1} (1)) \)
We know \( \tan^{-1} 1 = \frac{\pi}{4} \).
\( I = \frac{1}{2} \left( \tan^{-1} (e^2) - \frac{\pi}{4} \right) \)
(v) Let \( I = \int_1^2 3 x \sqrt{5-x^2} d x \)
Use substitution for this integral. Let \( t = 5-x^2 \).
Differentiate \( t \) with respect to \( x \): \( dt = -2x dx \).
So, \( x dx = -\frac{1}{2} dt \).
Change the limits. When \( x = 1 \), \( t = 5-1^2 = 4 \).
When \( x = 2 \), \( t = 5-2^2 = 5-4 = 1 \).
Substitute these into the integral:
\( I = \int_4^1 3 \sqrt{t} \left( -\frac{1}{2} dt \right) \)
\( I = -\frac{3}{2} \int_4^1 t^{1/2} dt \)
To make calculation easier, we can swap the limits and change the sign:
\( I = \frac{3}{2} \int_1^4 t^{1/2} dt \)
Integrate \( t^{1/2} \): \( \int t^{1/2} dt = \frac{t^{1/2+1}}{1/2+1} = \frac{t^{3/2}}{3/2} = \frac{2}{3} t^{3/2} \).
\( I = \frac{3}{2} \left[ \frac{2}{3} t^{3/2} \right]_1^4 \)
\( I = [t^{3/2}]_1^4 \)
Apply the limits:
\( I = 4^{3/2} - 1^{3/2} \)
\( I = (2^2)^{3/2} - 1 = 2^3 - 1 = 8 - 1 = 7 \)
(vi) Let \( I = \int_1^2 \frac{e^{-\frac{1}{x}}}{x^2} d x \)
Use substitution. Let \( t = -\frac{1}{x} \).
Differentiate \( t \) with respect to \( x \): \( dt = \frac{1}{x^2} dx \).
Change the limits. When \( x = 1 \), \( t = -\frac{1}{1} = -1 \).
When \( x = 2 \), \( t = -\frac{1}{2} \).
Substitute these into the integral:
\( I = \int_{-1}^{-1/2} e^t dt \)
Integrate \( e^t \): \( \int e^t dt = e^t \).
\( I = [e^t]_{-1}^{-1/2} \)
Apply the limits:
\( I = e^{-1/2} - e^{-1} \)
\( I = \frac{1}{\sqrt{e}} - \frac{1}{e} \)
(vii) Let \( I = \int_0^a x \sqrt{a^2-x^2} d x \)
Use substitution. Let \( t = a^2-x^2 \).
Differentiate \( t \) with respect to \( x \): \( dt = -2x dx \).
So, \( x dx = -\frac{1}{2} dt \).
Change the limits. When \( x = 0 \), \( t = a^2-0^2 = a^2 \).
When \( x = a \), \( t = a^2-a^2 = 0 \).
Substitute these into the integral:
\( I = \int_{a^2}^0 \sqrt{t} \left( -\frac{1}{2} dt \right) \)
\( I = -\frac{1}{2} \int_{a^2}^0 t^{1/2} dt \)
Swap the limits and change the sign:
\( I = \frac{1}{2} \int_0^{a^2} t^{1/2} dt \)
Integrate \( t^{1/2} \): \( \int t^{1/2} dt = \frac{t^{3/2}}{3/2} = \frac{2}{3} t^{3/2} \).
\( I = \frac{1}{2} \left[ \frac{2}{3} t^{3/2} \right]_0^{a^2} \)
\( I = \frac{1}{3} [t^{3/2}]_0^{a^2} \)
Apply the limits:
\( I = \frac{1}{3} ((a^2)^{3/2} - 0^{3/2}) \)
\( I = \frac{1}{3} (a^3 - 0) = \frac{a^3}{3} \)
(viii) Let \( I = \int_0^{(\pi / 2)^{1 / 3}} x^2 \sin x^3 d x \)
Use substitution. Let \( t = x^3 \).
Differentiate \( t \) with respect to \( x \): \( dt = 3x^2 dx \).
So, \( x^2 dx = \frac{1}{3} dt \).
Change the limits. When \( x = 0 \), \( t = 0^3 = 0 \).
When \( x = (\frac{\pi}{2})^{1/3} \), \( t = ((\frac{\pi}{2})^{1/3})^3 = \frac{\pi}{2} \).
Substitute these into the integral:
\( I = \int_0^{\pi/2} \sin t \left( \frac{1}{3} dt \right) \)
\( I = \frac{1}{3} \int_0^{\pi/2} \sin t dt \)
Integrate \( \sin t \): \( \int \sin t dt = -\cos t \).
\( I = \frac{1}{3} [-\cos t]_0^{\pi/2} \)
Apply the limits:
\( I = -\frac{1}{3} (\cos (\frac{\pi}{2}) - \cos 0) \)
We know \( \cos (\frac{\pi}{2}) = 0 \) and \( \cos 0 = 1 \).
\( I = -\frac{1}{3} (0 - 1) = -\frac{1}{3} (-1) = \frac{1}{3} \)
In simple words: For each problem, we changed the variable using substitution to make the integral easier to solve. We also updated the start and end points for the integral based on the new variable. Then, we solved the simpler integral and put in the new start and end numbers to get the final answer. This method is like translating the problem into a simpler language to find the solution.

🎯 Exam Tip: When using substitution for definite integrals, always remember to change the limits of integration according to the new variable; otherwise, your answer will be incorrect.

Question 2. Evaluate the following definite integrals:
(i) \( \int_0^{\pi / 2} \frac{\cos x}{1+\sin ^2 x} d x \)
(ii) \( \int_0^{\pi / 2} \frac{\sin x}{1+\cos ^2 x} d x \)
(iii) \( \int_0^{\pi / 3} \frac{\sec x \tan x}{1+\sec ^2 x} d x \)
(iv) \( \int_0^{\pi / 2} \frac{\sin x \cos x}{1+\sin ^4 x} d x \)
(v) \( \int_0^{\pi / 2} \frac{d x}{4 \sin ^2 x+5 \cos ^2 x} \)
(vi) \( \int_0^1 \frac{\tan ^{-1} x}{1+x^2} d x \)
(vii) \( \int_0^{\pi / 4} \frac{\tan ^3 x}{1+\cos 2 x} d x \)
(viii) \( \int_1^e \frac{\cos (\log x)}{x} d x \)
(ix) \( \int_0^{\pi / 2} \sqrt{\cos \theta} \cdot \sin ^3 \theta d \theta \)
(x) \( \int_0^{\pi / 4} 2 \tan ^3 x d x \)
(xi) \( \int_0^{\pi / 2} \frac{\sin x \cos x}{\cos ^2 x+3 \cos x+2} d x \)
(xii) \( \int_0^{\pi / 2} \frac{\cos x}{(1+\sin x)(2+\sin x)} d x \)
Answer:
(i) Let \( I = \int_0^{\pi / 2} \frac{\cos x d x}{1+\sin ^2 x} \)
Use substitution. Let \( t = \sin x \).
Differentiate \( t \) with respect to \( x \): \( dt = \cos x dx \).
Change the limits. When \( x = 0 \), \( t = \sin 0 = 0 \).
When \( x = \frac{\pi}{2} \), \( t = \sin \frac{\pi}{2} = 1 \).
Substitute into the integral:
\( I = \int_0^1 \frac{1}{1+t^2} dt \)
Integrate \( \frac{1}{1+t^2} \): \( \int \frac{1}{1+t^2} dt = \tan^{-1} t \).
\( I = [\tan^{-1} t]_0^1 \)
Apply the limits:
\( I = \tan^{-1} 1 - \tan^{-1} 0 \)
\( I = \frac{\pi}{4} - 0 = \frac{\pi}{4} \)
(ii) Let \( I = \int_0^{\pi / 2} \frac{\sin x d x}{1+\cos ^2 x} \)
Use substitution. Let \( t = \cos x \).
Differentiate \( t \) with respect to \( x \): \( dt = -\sin x dx \).
So, \( \sin x dx = -dt \).
Change the limits. When \( x = 0 \), \( t = \cos 0 = 1 \).
When \( x = \frac{\pi}{2} \), \( t = \cos \frac{\pi}{2} = 0 \).
Substitute into the integral:
\( I = \int_1^0 \frac{-dt}{1+t^2} \)
Swap the limits and change the sign:
\( I = \int_0^1 \frac{dt}{1+t^2} \)
Integrate \( \frac{1}{1+t^2} \): \( \int \frac{1}{1+t^2} dt = \tan^{-1} t \).
\( I = [\tan^{-1} t]_0^1 \)
Apply the limits:
\( I = \tan^{-1} 1 - \tan^{-1} 0 \)
\( I = \frac{\pi}{4} - 0 = \frac{\pi}{4} \)
(iii) Let \( I = \int_0^{\pi / 3} \frac{\sec x \tan x d x}{1+\sec ^2 x} \)
Use substitution. Let \( t = \sec x \).
Differentiate \( t \) with respect to \( x \): \( dt = \sec x \tan x dx \).
Change the limits. When \( x = 0 \), \( t = \sec 0 = 1 \).
When \( x = \frac{\pi}{3} \), \( t = \sec \frac{\pi}{3} = 2 \).
Substitute into the integral:
\( I = \int_1^2 \frac{dt}{1+t^2} \)
Integrate \( \frac{1}{1+t^2} \): \( \int \frac{1}{1+t^2} dt = \tan^{-1} t \).
\( I = [\tan^{-1} t]_1^2 \)
Apply the limits:
\( I = \tan^{-1} 2 - \tan^{-1} 1 \)
We know \( \tan^{-1} 1 = \frac{\pi}{4} \).
\( I = \tan^{-1} 2 - \frac{\pi}{4} \)
(iv) Let \( I = \int_0^{\pi / 2} \frac{\sin x \cos x d x}{1+\sin ^4 x} \)
Use substitution. Let \( t = \sin^2 x \).
Differentiate \( t \) with respect to \( x \): \( dt = 2 \sin x \cos x dx \).
So, \( \sin x \cos x dx = \frac{1}{2} dt \). Also, \( \sin^4 x = ( \sin^2 x )^2 = t^2 \).
Change the limits. When \( x = 0 \), \( t = \sin^2 0 = 0 \).
When \( x = \frac{\pi}{2} \), \( t = \sin^2 \frac{\pi}{2} = 1 \).
Substitute into the integral:
\( I = \int_0^1 \frac{1}{1+t^2} \left( \frac{1}{2} dt \right) \)
\( I = \frac{1}{2} \int_0^1 \frac{1}{1+t^2} dt \)
Integrate \( \frac{1}{1+t^2} \): \( \int \frac{1}{1+t^2} dt = \tan^{-1} t \).
\( I = \frac{1}{2} [\tan^{-1} t]_0^1 \)
Apply the limits:
\( I = \frac{1}{2} (\tan^{-1} 1 - \tan^{-1} 0) \)
\( I = \frac{1}{2} \left( \frac{\pi}{4} - 0 \right) = \frac{\pi}{8} \)
(v) Let \( I = \int_0^{\pi / 2} \frac{d x}{4 \sin ^2 x+5 \cos ^2 x} \)
Divide both the numerator and denominator by \( \cos^2 x \):
\( I = \int_0^{\pi / 2} \frac{\frac{1}{\cos^2 x}}{\frac{4 \sin^2 x}{\cos^2 x}+\frac{5 \cos^2 x}{\cos^2 x}} d x \)
\( I = \int_0^{\pi / 2} \frac{\sec^2 x}{4 \tan^2 x+5} d x \)
Now, use substitution. Let \( t = \tan x \).
Differentiate \( t \) with respect to \( x \): \( dt = \sec^2 x dx \).
Change the limits. When \( x = 0 \), \( t = \tan 0 = 0 \).
When \( x = \frac{\pi}{2} \), \( t = \tan \frac{\pi}{2} = \infty \).
Substitute into the integral:
\( I = \int_0^{\infty} \frac{dt}{4t^2+5} \)
Factor out 4 from the denominator:
\( I = \frac{1}{4} \int_0^{\infty} \frac{dt}{t^2+\frac{5}{4}} \)
Rewrite the constant term: \( \frac{5}{4} = \left(\frac{\sqrt{5}}{2}\right)^2 \).
\( I = \frac{1}{4} \int_0^{\infty} \frac{dt}{t^2+\left(\frac{\sqrt{5}}{2}\right)^2} \)
This is a standard integral form: \( \int \frac{1}{x^2+a^2} dx = \frac{1}{a} \tan^{-1} \frac{x}{a} \). Here \( a = \frac{\sqrt{5}}{2} \).
\( I = \frac{1}{4} \left[ \frac{1}{\frac{\sqrt{5}}{2}} \tan^{-1} \frac{t}{\frac{\sqrt{5}}{2}} \right]_0^{\infty} \)
\( I = \frac{1}{4} \cdot \frac{2}{\sqrt{5}} \left[ \tan^{-1} \frac{2t}{\sqrt{5}} \right]_0^{\infty} \)
\( I = \frac{1}{2\sqrt{5}} (\tan^{-1} \infty - \tan^{-1} 0) \)
We know \( \tan^{-1} \infty = \frac{\pi}{2} \) and \( \tan^{-1} 0 = 0 \).
\( I = \frac{1}{2\sqrt{5}} \left( \frac{\pi}{2} - 0 \right) = \frac{\pi}{4\sqrt{5}} \)
(vi) Let \( I = \int_0^1 \frac{\tan ^{-1} x}{1+x^2} d x \)
Use substitution. Let \( t = \tan^{-1} x \).
Differentiate \( t \) with respect to \( x \): \( dt = \frac{1}{1+x^2} dx \).
Change the limits. When \( x = 0 \), \( t = \tan^{-1} 0 = 0 \).
When \( x = 1 \), \( t = \tan^{-1} 1 = \frac{\pi}{4} \).
Substitute into the integral:
\( I = \int_0^{\pi/4} t dt \)
Integrate \( t \): \( \int t dt = \frac{t^2}{2} \).
\( I = \left[ \frac{t^2}{2} \right]_0^{\pi/4} \)
Apply the limits:
\( I = \frac{1}{2} \left( \left(\frac{\pi}{4}\right)^2 - 0^2 \right) \)
\( I = \frac{1}{2} \left( \frac{\pi^2}{16} \right) = \frac{\pi^2}{32} \)
(vii) Let \( I = \int_0^{\pi / 4} \frac{\tan ^3 x}{1+\cos 2 x} d x \)
First, use the trigonometric identity \( 1+\cos 2x = 2 \cos^2 x \).
\( I = \int_0^{\pi / 4} \frac{\tan ^3 x}{2 \cos^2 x} d x \)
Rewrite \( \frac{1}{\cos^2 x} \) as \( \sec^2 x \):
\( I = \frac{1}{2} \int_0^{\pi / 4} \tan ^3 x \sec^2 x d x \)
Use substitution. Let \( t = \tan x \).
Differentiate \( t \) with respect to \( x \): \( dt = \sec^2 x dx \).
Change the limits. When \( x = 0 \), \( t = \tan 0 = 0 \).
When \( x = \frac{\pi}{4} \), \( t = \tan \frac{\pi}{4} = 1 \).
Substitute into the integral:
\( I = \frac{1}{2} \int_0^1 t^3 dt \)
Integrate \( t^3 \): \( \int t^3 dt = \frac{t^4}{4} \).
\( I = \frac{1}{2} \left[ \frac{t^4}{4} \right]_0^1 \)
Apply the limits:
\( I = \frac{1}{8} (1^4 - 0^4) = \frac{1}{8} (1 - 0) = \frac{1}{8} \)
(viii) Let \( I = \int_1^e \frac{\cos (\log x)}{x} d x \)
Use substitution. Let \( t = \log x \).
Differentiate \( t \) with respect to \( x \): \( dt = \frac{1}{x} dx \).
Change the limits. When \( x = 1 \), \( t = \log 1 = 0 \).
When \( x = e \), \( t = \log e = 1 \).
Substitute into the integral:
\( I = \int_0^1 \cos t dt \)
Integrate \( \cos t \): \( \int \cos t dt = \sin t \).
\( I = [\sin t]_0^1 \)
Apply the limits:
\( I = \sin 1 - \sin 0 \)
Since \( \sin 0 = 0 \):
\( I = \sin 1 \)
(ix) Let \( I = \int_0^{\pi / 2} \sqrt{\cos \theta} \cdot \sin ^3 \theta d \theta \)
Rewrite \( \sin^3 \theta \) as \( \sin^2 \theta \cdot \sin \theta \), then use \( \sin^2 \theta = 1-\cos^2 \theta \):
\( I = \int_0^{\pi / 2} \sqrt{\cos \theta} (1-\cos^2 \theta) \sin \theta d \theta \)
Use substitution. Let \( t = \cos \theta \).
Differentiate \( t \) with respect to \( \theta \): \( dt = -\sin \theta d \theta \).
So, \( \sin \theta d \theta = -dt \).
Change the limits. When \( \theta = 0 \), \( t = \cos 0 = 1 \).
When \( \theta = \frac{\pi}{2} \), \( t = \cos \frac{\pi}{2} = 0 \).
Substitute into the integral:
\( I = \int_1^0 \sqrt{t} (1-t^2) (-dt) \)
\( I = -\int_1^0 (t^{1/2} - t^{5/2}) dt \)
Swap the limits and change the sign:
\( I = \int_0^1 (t^{1/2} - t^{5/2}) dt \)
Integrate each term:
\( \int t^{1/2} dt = \frac{t^{3/2}}{3/2} = \frac{2}{3} t^{3/2} \)
\( \int t^{5/2} dt = \frac{t^{7/2}}{7/2} = \frac{2}{7} t^{7/2} \)
\( I = \left[ \frac{2}{3} t^{3/2} - \frac{2}{7} t^{7/2} \right]_0^1 \)
Apply the limits:
\( I = \left( \frac{2}{3} (1)^{3/2} - \frac{2}{7} (1)^{7/2} \right) - \left( \frac{2}{3} (0)^{3/2} - \frac{2}{7} (0)^{7/2} \right) \)
\( I = \left( \frac{2}{3} - \frac{2}{7} \right) - (0 - 0) \)
\( I = \frac{14 - 6}{21} = \frac{8}{21} \)
(x) Let \( I = \int_0^{\pi / 4} 2 \tan ^3 x d x \)
Rewrite \( \tan^3 x \) as \( \tan x \cdot \tan^2 x \), then use \( \tan^2 x = \sec^2 x - 1 \):
\( I = 2 \int_0^{\pi / 4} \tan x (\sec^2 x - 1) d x \)
\( I = 2 \int_0^{\pi / 4} \tan x \sec^2 x d x - 2 \int_0^{\pi / 4} \tan x d x \)
For the first integral, let \( u = \tan x \), so \( du = \sec^2 x dx \).
When \( x = 0 \), \( u = 0 \). When \( x = \frac{\pi}{4} \), \( u = 1 \).
\( 2 \int_0^1 u du = 2 \left[ \frac{u^2}{2} \right]_0^1 = 2 \left( \frac{1^2}{2} - \frac{0^2}{2} \right) = 2 \left( \frac{1}{2} \right) = 1 \).
For the second integral, \( \int \tan x dx = -\log |\cos x| \).
\( -2 \int_0^{\pi / 4} \tan x d x = -2 [-\log |\cos x|]_0^{\pi / 4} \)
\( = 2 [\log |\cos x|]_0^{\pi / 4} \)
\( = 2 (\log |\cos \frac{\pi}{4}| - \log |\cos 0|) \)
\( = 2 (\log |\frac{1}{\sqrt{2}}| - \log |1|) \)
\( = 2 (-\frac{1}{2} \log 2 - 0) = -\log 2 \).
Combining both parts:
\( I = 1 - \log 2 \)
(xi) Let \( I = \int_0^{\pi / 2} \frac{\sin x \cos x}{\cos ^2 x+3 \cos x+2} d x \)
Use substitution. Let \( t = \cos x \).
Differentiate \( t \) with respect to \( x \): \( dt = -\sin x dx \).
So, \( \sin x dx = -dt \). And \( \cos x \) becomes \( t \).
Change the limits. When \( x = 0 \), \( t = \cos 0 = 1 \).
When \( x = \frac{\pi}{2} \), \( t = \cos \frac{\pi}{2} = 0 \).
Substitute into the integral:
\( I = \int_1^0 \frac{t (-dt)}{t^2+3t+2} \)
\( I = -\int_1^0 \frac{t}{t^2+3t+2} dt \)
Swap the limits and change the sign:
\( I = \int_0^1 \frac{t}{t^2+3t+2} dt \)
Factor the denominator: \( t^2+3t+2 = (t+1)(t+2) \).
Use partial fractions: \( \frac{t}{(t+1)(t+2)} = \frac{A}{t+1} + \frac{B}{t+2} \).
Multiplying by \( (t+1)(t+2) \): \( t = A(t+2) + B(t+1) \).
If \( t = -1 \), then \( -1 = A(-1+2) + B(0) \implies -1 = A \).
If \( t = -2 \), then \( -2 = A(0) + B(-2+1) \implies -2 = -B \implies B = 2 \).
So, \( \frac{t}{(t+1)(t+2)} = \frac{-1}{t+1} + \frac{2}{t+2} \).
\( I = \int_0^1 \left( \frac{-1}{t+1} + \frac{2}{t+2} \right) dt \)
\( I = [-\log |t+1| + 2\log |t+2|]_0^1 \)
Apply the limits:
\( I = (-\log |1+1| + 2\log |1+2|) - (-\log |0+1| + 2\log |0+2|) \)
\( I = (-\log 2 + 2\log 3) - (-\log 1 + 2\log 2) \)
Since \( \log 1 = 0 \):
\( I = -\log 2 + 2\log 3 - 2\log 2 \)
\( I = 2\log 3 - 3\log 2 \)
Using logarithm properties: \( \log a^n = n\log a \) and \( \log a - \log b = \log(a/b) \):
\( I = \log 3^2 - \log 2^3 = \log 9 - \log 8 = \log \left( \frac{9}{8} \right) \)
(xii) Let \( I = \int_0^{\pi / 2} \frac{\cos x d x}{(1+\sin x)(2+\sin x)} \)
Use substitution. Let \( t = \sin x \).
Differentiate \( t \) with respect to \( x \): \( dt = \cos x dx \).
Change the limits. When \( x = 0 \), \( t = \sin 0 = 0 \).
When \( x = \frac{\pi}{2} \), \( t = \sin \frac{\pi}{2} = 1 \).
Substitute into the integral:
\( I = \int_0^1 \frac{dt}{(1+t)(2+t)} \)
Use partial fractions: \( \frac{1}{(1+t)(2+t)} = \frac{A}{1+t} + \frac{B}{2+t} \).
Multiplying by \( (1+t)(2+t) \): \( 1 = A(2+t) + B(1+t) \).
If \( t = -1 \), then \( 1 = A(2-1) + B(0) \implies A = 1 \).
If \( t = -2 \), then \( 1 = A(0) + B(1-2) \implies 1 = -B \implies B = -1 \).
So, \( \frac{1}{(1+t)(2+t)} = \frac{1}{1+t} - \frac{1}{2+t} \).
\( I = \int_0^1 \left( \frac{1}{1+t} - \frac{1}{2+t} \right) dt \)
\( I = [\log |1+t| - \log |2+t|]_0^1 \)
\( I = \left[ \log \left| \frac{1+t}{2+t} \right| \right]_0^1 \)
Apply the limits:
\( I = \log \left| \frac{1+1}{2+1} \right| - \log \left| \frac{1+0}{2+0} \right| \)
\( I = \log \left( \frac{2}{3} \right) - \log \left( \frac{1}{2} \right) \)
Using logarithm properties:
\( I = \log \left( \frac{2/3}{1/2} \right) = \log \left( \frac{2}{3} \times 2 \right) = \log \left( \frac{4}{3} \right) \)
In simple words: For each problem, we made the integral simpler by changing the variable and adjusting the start and end points. Then, we solved the new integral. Sometimes, we used tricks like dividing by \( \cos^2 x \) or breaking fractions into smaller ones to make the integration easier. Finally, we put the new limits back into the solved integral to get the answer.

🎯 Exam Tip: When dealing with integrals involving trigonometric functions, remember to look for opportunities to use identities (like \( 1+\cos 2x = 2\cos^2 x \)) or divide by terms like \( \cos^2 x \) to simplify the expression into a form suitable for substitution.

Question 3. Evaluate the following definite integrals:
(i) \( \int_0^1 \sqrt{\frac{1-x}{1+x}} d x \)
(ii) \( \int_0^1 x \sqrt{\frac{1-x}{1+x}} d x \)
(iii) \( \int_{-a}^a \sqrt{\frac{a-x}{a+x}} d x \)
(iv) \( \int_0^2 \frac{6 x+3}{x^2+4} d x \)
Answer:
(i) Let \( I = \int_0^1 \sqrt{\frac{1-x}{1+x}} d x \)
Use substitution. Let \( x = \cos \theta \).
Differentiate \( x \) with respect to \( \theta \): \( dx = -\sin \theta d \theta \).
Change the limits. When \( x = 0 \), \( \cos \theta = 0 \implies \theta = \frac{\pi}{2} \).
When \( x = 1 \), \( \cos \theta = 1 \implies \theta = 0 \).
Substitute into the integral:
\( I = \int_{\pi/2}^0 \sqrt{\frac{1-\cos \theta}{1+\cos \theta}} (-\sin \theta d \theta) \)
Use half-angle identities: \( 1-\cos \theta = 2\sin^2 (\theta/2) \) and \( 1+\cos \theta = 2\cos^2 (\theta/2) \).
\( \sqrt{\frac{1-\cos \theta}{1+\cos \theta}} = \sqrt{\frac{2\sin^2 (\theta/2)}{2\cos^2 (\theta/2)}} = \sqrt{\tan^2 (\theta/2)} = \tan (\theta/2) \).
Also, \( \sin \theta = 2\sin (\theta/2) \cos (\theta/2) \).
\( I = \int_{\pi/2}^0 \tan (\theta/2) (-2\sin (\theta/2) \cos (\theta/2)) d \theta \)
\( I = \int_{\pi/2}^0 \frac{\sin (\theta/2)}{\cos (\theta/2)} (-2\sin (\theta/2) \cos (\theta/2)) d \theta \)
\( I = \int_{\pi/2}^0 -2\sin^2 (\theta/2) d \theta \)
Use the identity \( 2\sin^2 (\theta/2) = 1-\cos \theta \):
\( I = \int_{\pi/2}^0 -(1-\cos \theta) d \theta = \int_{\pi/2}^0 (\cos \theta - 1) d \theta \)
Integrate term by term:
\( I = [\sin \theta - \theta]_{\pi/2}^0 \)
Apply the limits:
\( I = (\sin 0 - 0) - (\sin \frac{\pi}{2} - \frac{\pi}{2}) \)
\( I = (0 - 0) - (1 - \frac{\pi}{2}) \)
\( I = -1 + \frac{\pi}{2} = \frac{\pi}{2} - 1 \)
(ii) Let \( I = \int_0^1 x \sqrt{\frac{1-x^2}{1+x^2}} d x \)
Use substitution. Let \( x^2 = \cos \theta \).
Differentiate with respect to \( x \): \( 2x dx = -\sin \theta d \theta \).
So, \( x dx = -\frac{1}{2} \sin \theta d \theta \).
Change the limits. When \( x = 0 \), \( \cos \theta = 0 \implies \theta = \frac{\pi}{2} \).
When \( x = 1 \), \( \cos \theta = 1 \implies \theta = 0 \).
Substitute into the integral:
\( I = \int_{\pi/2}^0 \sqrt{\frac{1-\cos \theta}{1+\cos \theta}} \left( -\frac{1}{2} \sin \theta d \theta \right) \)
Using identities \( \sqrt{\frac{1-\cos \theta}{1+\cos \theta}} = \tan (\theta/2) \) and \( \sin \theta = 2\sin (\theta/2) \cos (\theta/2) \):
\( I = \int_{\pi/2}^0 \tan (\theta/2) \left( -\frac{1}{2} \cdot 2\sin (\theta/2) \cos (\theta/2) \right) d \theta \)
\( I = \int_{\pi/2}^0 \frac{\sin (\theta/2)}{\cos (\theta/2)} (-\sin (\theta/2) \cos (\theta/2)) d \theta \)
\( I = \int_{\pi/2}^0 -\sin^2 (\theta/2) d \theta \)
Use the identity \( \sin^2 (\theta/2) = \frac{1-\cos \theta}{2} \):
\( I = \int_{\pi/2}^0 -\left( \frac{1-\cos \theta}{2} \right) d \theta \)
\( I = -\frac{1}{2} [\theta - \sin \theta]_{\pi/2}^0 \)
Apply the limits:
\( I = -\frac{1}{2} [(\theta - \sin \theta)]_{\pi/2}^0 \)
\( I = -\frac{1}{2} [(0 - \sin 0) - (\frac{\pi}{2} - \sin \frac{\pi}{2})] \)
\( I = -\frac{1}{2} [0 - (\frac{\pi}{2} - 1)] \)
\( I = -\frac{1}{2} [-\frac{\pi}{2} + 1] = \frac{\pi}{4} - \frac{1}{2} \)
(iii) Let \( I = \int_{-a}^a \sqrt{\frac{a-x}{a+x}} d x \)
Rewrite the integrand by multiplying the numerator and denominator inside the square root by \( (a-x) \):
\( I = \int_{-a}^a \sqrt{\frac{(a-x)(a-x)}{(a+x)(a-x)}} d x = \int_{-a}^a \sqrt{\frac{(a-x)^2}{a^2-x^2}} d x \)
\( I = \int_{-a}^a \frac{|a-x|}{\sqrt{a^2-x^2}} d x \)
Since \( x \) ranges from \( -a \) to \( a \), \( a-x \) is always positive, so \( |a-x| = a-x \).
\( I = \int_{-a}^a \frac{a-x}{\sqrt{a^2-x^2}} d x \)
Split the integral into two parts:
\( I = \int_{-a}^a \frac{a}{\sqrt{a^2-x^2}} d x - \int_{-a}^a \frac{x}{\sqrt{a^2-x^2}} d x \)
For the first part, \( \int \frac{1}{\sqrt{a^2-x^2}} dx = \sin^{-1} \frac{x}{a} \):
\( a \left[ \sin^{-1} \frac{x}{a} \right]_{-a}^a = a \left( \sin^{-1} \frac{a}{a} - \sin^{-1} \frac{-a}{a} \right) \)
\( = a (\sin^{-1} 1 - \sin^{-1} (-1)) = a \left( \frac{\pi}{2} - (-\frac{\pi}{2}) \right) = a \left( \frac{\pi}{2} + \frac{\pi}{2} \right) = a\pi \).
For the second part, let \( u = a^2-x^2 \). Then \( du = -2x dx \implies x dx = -\frac{1}{2} du \).
When \( x = -a \), \( u = a^2-(-a)^2 = 0 \). When \( x = a \), \( u = a^2-a^2 = 0 \).
So, \( \int_{-a}^a \frac{x}{\sqrt{a^2-x^2}} d x = \int_0^0 \frac{-\frac{1}{2} du}{\sqrt{u}} = 0 \).
Combining both parts:
\( I = a\pi - 0 = a\pi \)
(iv) Let \( I = \int_0^2 \frac{6 x+3}{x^2+4} d x \)
Split the integral into two parts:
\( I = \int_0^2 \frac{6x}{x^2+4} d x + \int_0^2 \frac{3}{x^2+4} d x \)
For the first integral, let \( u = x^2+4 \). Then \( du = 2x dx \).
So, \( 6x dx = 3 \cdot (2x dx) = 3 du \).
Change limits: When \( x = 0 \), \( u = 0^2+4 = 4 \). When \( x = 2 \), \( u = 2^2+4 = 8 \).
\( \int_4^8 \frac{3 du}{u} = 3[\log |u|]_4^8 = 3(\log 8 - \log 4) = 3 \log \left( \frac{8}{4} \right) = 3 \log 2 \).
For the second integral, this is a standard form \( \int \frac{1}{x^2+a^2} dx = \frac{1}{a} \tan^{-1} \frac{x}{a} \). Here \( a = \sqrt{4} = 2 \).
\( 3 \int_0^2 \frac{d x}{x^2+2^2} = 3 \left[ \frac{1}{2} \tan^{-1} \frac{x}{2} \right]_0^2 \)
\( = \frac{3}{2} \left( \tan^{-1} \frac{2}{2} - \tan^{-1} \frac{0}{2} \right) \)
\( = \frac{3}{2} (\tan^{-1} 1 - \tan^{-1} 0) = \frac{3}{2} \left( \frac{\pi}{4} - 0 \right) = \frac{3\pi}{8} \).
Combining both parts:
\( I = 3 \log 2 + \frac{3\pi}{8} \)
In simple words: These problems need different methods like changing variables or splitting the integral into smaller, easier parts. Sometimes, we use special trigonometric rules to simplify the expression first. Always remember to change the limits of the integral when you change the variable. For integrals with square roots in the denominator, look for ways to use \( \sin^{-1} \) or \( \tan^{-1} \) forms.

🎯 Exam Tip: For integrals like \( \int_{-a}^a f(x) dx \), check if \( f(x) \) is an odd function (result is 0) or an even function (result is \( 2 \int_0^a f(x) dx \)). This can save time. Here, \( \frac{a}{\sqrt{a^2-x^2}} \) is even, and \( \frac{x}{\sqrt{a^2-x^2}} \) is odd.

Question 4. Evaluate the following definite integrals:
(i) \( \int_0^{\pi / 2} \frac{d x}{3+2 \cos x} \)
(ii) \( \int_0^{\pi / 2} \frac{d x}{7+4 \sin x} \)
(iii) \( \int_0^{\pi / 2} \frac{\sin 2 x}{\sin ^4 x+\cos ^4 x} d x \)
Answer:
(i) Let \( I = \int_0^{\pi / 2} \frac{d x}{3+2 \cos x} \)
Use the substitution \( t = \tan \frac{x}{2} \).
Then \( dx = \frac{2 dt}{1+t^2} \) and \( \cos x = \frac{1-t^2}{1+t^2} \).
Change the limits. When \( x = 0 \), \( t = \tan 0 = 0 \).
When \( x = \frac{\pi}{2} \), \( t = \tan \frac{\pi}{4} = 1 \).
Substitute into the integral:
\( I = \int_0^1 \frac{\frac{2 dt}{1+t^2}}{3+2\left(\frac{1-t^2}{1+t^2}\right)} \)
Simplify the denominator:
\( 3+2\left(\frac{1-t^2}{1+t^2}\right) = \frac{3(1+t^2)+2(1-t^2)}{1+t^2} = \frac{3+3t^2+2-2t^2}{1+t^2} = \frac{5+t^2}{1+t^2} \).
So, \( I = \int_0^1 \frac{\frac{2 dt}{1+t^2}}{\frac{5+t^2}{1+t^2}} = \int_0^1 \frac{2 dt}{t^2+5} \)
\( I = 2 \int_0^1 \frac{dt}{t^2+(\sqrt{5})^2} \)
This is a standard integral form: \( \int \frac{1}{x^2+a^2} dx = \frac{1}{a} \tan^{-1} \frac{x}{a} \). Here \( a = \sqrt{5} \).
\( I = 2 \left[ \frac{1}{\sqrt{5}} \tan^{-1} \frac{t}{\sqrt{5}} \right]_0^1 \)
Apply the limits:
\( I = \frac{2}{\sqrt{5}} \left( \tan^{-1} \frac{1}{\sqrt{5}} - \tan^{-1} 0 \right) \)
\( I = \frac{2}{\sqrt{5}} \tan^{-1} \frac{1}{\sqrt{5}} \)
(ii) Let \( I = \int_0^{\pi / 2} \frac{d x}{7+4 \sin x} \)
Use the substitution \( t = \tan \frac{x}{2} \).
Then \( dx = \frac{2 dt}{1+t^2} \) and \( \sin x = \frac{2t}{1+t^2} \).
Change the limits. When \( x = 0 \), \( t = \tan 0 = 0 \).
When \( x = \frac{\pi}{2} \), \( t = \tan \frac{\pi}{4} = 1 \).
Substitute into the integral:
\( I = \int_0^1 \frac{\frac{2 dt}{1+t^2}}{7+4\left(\frac{2t}{1+t^2}\right)} \)
Simplify the denominator:
\( 7+4\left(\frac{2t}{1+t^2}\right) = \frac{7(1+t^2)+8t}{1+t^2} = \frac{7t^2+8t+7}{1+t^2} \).
So, \( I = \int_0^1 \frac{\frac{2 dt}{1+t^2}}{\frac{7t^2+8t+7}{1+t^2}} = \int_0^1 \frac{2 dt}{7t^2+8t+7} \)
Factor out 7 from the denominator:
\( I = \frac{2}{7} \int_0^1 \frac{dt}{t^2+\frac{8}{7}t+1} \)
Complete the square in the denominator: \( t^2+\frac{8}{7}t+1 = (t+\frac{4}{7})^2 - (\frac{4}{7})^2 + 1 = (t+\frac{4}{7})^2 + \frac{33}{49} = (t+\frac{4}{7})^2 + \left(\frac{\sqrt{33}}{7}\right)^2 \).
\( I = \frac{2}{7} \int_0^1 \frac{dt}{\left(t+\frac{4}{7}\right)^2 + \left(\frac{\sqrt{33}}{7}\right)^2} \)
This is a standard integral form. Here, \( X = t+\frac{4}{7} \) and \( A = \frac{\sqrt{33}}{7} \).
\( I = \frac{2}{7} \left[ \frac{1}{\frac{\sqrt{33}}{7}} \tan^{-1} \frac{t+\frac{4}{7}}{\frac{\sqrt{33}}{7}} \right]_0^1 \)
\( I = \frac{2}{7} \cdot \frac{7}{\sqrt{33}} \left[ \tan^{-1} \frac{7t+4}{\sqrt{33}} \right]_0^1 \)
\( I = \frac{2}{\sqrt{33}} \left( \tan^{-1} \frac{7(1)+4}{\sqrt{33}} - \tan^{-1} \frac{7(0)+4}{\sqrt{33}} \right) \)
\( I = \frac{2}{\sqrt{33}} \left( \tan^{-1} \frac{11}{\sqrt{33}} - \tan^{-1} \frac{4}{\sqrt{33}} \right) \)
(iii) Let \( I = \int_0^{\pi / 2} \frac{\sin 2 x}{\sin ^4 x+\cos ^4 x} d x \)
Use the identity \( \sin 2x = 2 \sin x \cos x \).
Divide both the numerator and denominator by \( \cos^4 x \):
\( I = \int_0^{\pi / 2} \frac{\frac{2 \sin x \cos x}{\cos^4 x}}{\frac{\sin^4 x}{\cos^4 x}+\frac{\cos^4 x}{\cos^4 x}} d x \)
\( I = \int_0^{\pi / 2} \frac{2 \tan x \sec^2 x}{\tan^4 x+1} d x \)
Use substitution. Let \( t = \tan^2 x \).
Differentiate \( t \) with respect to \( x \): \( dt = 2 \tan x \sec^2 x dx \).
Change the limits. When \( x = 0 \), \( t = \tan^2 0 = 0 \).
When \( x = \frac{\pi}{2} \), \( t = \tan^2 \frac{\pi}{2} = \infty \).
Substitute into the integral:
\( I = \int_0^{\infty} \frac{dt}{t^2+1} \)
This is a standard integral form: \( \int \frac{1}{x^2+a^2} dx = \frac{1}{a} \tan^{-1} \frac{x}{a} \). Here \( a = 1 \).
\( I = [\tan^{-1} t]_0^{\infty} \)
Apply the limits:
\( I = \tan^{-1} \infty - \tan^{-1} 0 \)
\( I = \frac{\pi}{2} - 0 = \frac{\pi}{2} \)
In simple words: For these integrals with \( \sin x \) and \( \cos x \) in the denominator, a common trick is to change everything to \( \tan(\frac{x}{2}) \) and \( \sec^2(\frac{x}{2}) \), or to divide by \( \cos^2 x \) or \( \cos^4 x \). This helps turn the integral into a simpler form that can be solved using standard formulas like for \( \tan^{-1} \).

🎯 Exam Tip: For integrals of the form \( \int \frac{dx}{a+b\sin x} \) or \( \int \frac{dx}{a+b\cos x} \), the substitution \( t = \tan \frac{x}{2} \) is very powerful. Remember the identities: \( \sin x = \frac{2t}{1+t^2} \), \( \cos x = \frac{1-t^2}{1+t^2} \), and \( dx = \frac{2dt}{1+t^2} \).

Question 5. Evaluate the following definite integrals:
(i) \( \int_0^{\sqrt{2}} \sqrt{2-x^2} d x \)
(ii) \( \int_0^3 \sqrt{9-x^2} d x \)
(iii) \( \int_0^a \frac{x^4}{\sqrt{a^2-x^2}} d x \)
(iv) \( \int_8^{15} \frac{d x}{(x-3) \sqrt{x+1}} \)
(v) \( \int_{1 / 3}^1 \frac{\left(x-x^3\right)^{1 / 3}}{x^4} d x \)
Answer:
(i) Let \( I = \int_0^{\sqrt{2}} \sqrt{2-x^2} d x \)
Use substitution. Let \( x = \sqrt{2} \sin \theta \).
Differentiate \( x \) with respect to \( \theta \): \( dx = \sqrt{2} \cos \theta d \theta \).
Change the limits. When \( x = 0 \), \( \sqrt{2} \sin \theta = 0 \implies \sin \theta = 0 \implies \theta = 0 \).
When \( x = \sqrt{2} \), \( \sqrt{2} \sin \theta = \sqrt{2} \implies \sin \theta = 1 \implies \theta = \frac{\pi}{2} \).
Substitute into the integral:
\( I = \int_0^{\pi/2} \sqrt{2-(\sqrt{2}\sin \theta)^2} (\sqrt{2}\cos \theta d \theta) \)
\( I = \int_0^{\pi/2} \sqrt{2-2\sin^2 \theta} (\sqrt{2}\cos \theta d \theta) \)
\( I = \int_0^{\pi/2} \sqrt{2(1-\sin^2 \theta)} (\sqrt{2}\cos \theta d \theta) \)
\( I = \int_0^{\pi/2} \sqrt{2\cos^2 \theta} (\sqrt{2}\cos \theta d \theta) \)
\( I = \int_0^{\pi/2} (\sqrt{2}\cos \theta) (\sqrt{2}\cos \theta d \theta) \)
\( I = \int_0^{\pi/2} 2\cos^2 \theta d \theta \)
Use the identity \( 2\cos^2 \theta = 1+\cos 2\theta \):
\( I = \int_0^{\pi/2} (1+\cos 2\theta) d \theta \)
Integrate term by term:
\( I = \left[ \theta + \frac{\sin 2\theta}{2} \right]_0^{\pi/2} \)
Apply the limits:
\( I = \left( \frac{\pi}{2} + \frac{\sin (2 \cdot \frac{\pi}{2})}{2} \right) - \left( 0 + \frac{\sin (2 \cdot 0)}{2} \right) \)
\( I = \left( \frac{\pi}{2} + \frac{\sin \pi}{2} \right) - (0 + \frac{\sin 0}{2}) \)
Since \( \sin \pi = 0 \) and \( \sin 0 = 0 \):
\( I = \left( \frac{\pi}{2} + 0 \right) - (0 + 0) = \frac{\pi}{2} \)
(ii) Let \( I = \int_0^3 \sqrt{9-x^2} d x \)
This integral is of the form \( \int \sqrt{a^2-x^2} dx \), where \( a=3 \).
The direct formula for this is \( \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a} \).
So, \( I = \left[ \frac{x}{2}\sqrt{9-x^2} + \frac{9}{2}\sin^{-1}\frac{x}{3} \right]_0^3 \)
Apply the limits:
\( I = \left( \frac{3}{2}\sqrt{9-3^2} + \frac{9}{2}\sin^{-1}\frac{3}{3} \right) - \left( \frac{0}{2}\sqrt{9-0^2} + \frac{9}{2}\sin^{-1}\frac{0}{3} \right) \)
\( I = \left( \frac{3}{2}\sqrt{0} + \frac{9}{2}\sin^{-1} 1 \right) - \left( 0 + \frac{9}{2}\sin^{-1} 0 \right) \)
\( I = \left( 0 + \frac{9}{2} \cdot \frac{\pi}{2} \right) - (0 + 0) \)
\( I = \frac{9\pi}{4} \)
(iii) Let \( I = \int_0^a \frac{x^4}{\sqrt{a^2-x^2}} d x \)
Use substitution. Let \( x = a \sin \theta \).
Differentiate \( x \) with respect to \( \theta \): \( dx = a \cos \theta d \theta \).
Change the limits. When \( x = 0 \), \( a \sin \theta = 0 \implies \sin \theta = 0 \implies \theta = 0 \).
When \( x = a \), \( a \sin \theta = a \implies \sin \theta = 1 \implies \theta = \frac{\pi}{2} \).
Substitute into the integral:
\( I = \int_0^{\pi/2} \frac{(a\sin \theta)^4}{\sqrt{a^2-(a\sin \theta)^2}} (a\cos \theta d \theta) \)
\( I = \int_0^{\pi/2} \frac{a^4\sin^4 \theta}{\sqrt{a^2(1-\sin^2 \theta)}} (a\cos \theta d \theta) \)
\( I = \int_0^{\pi/2} \frac{a^4\sin^4 \theta}{\sqrt{a^2\cos^2 \theta}} (a\cos \theta d \theta) \)
\( I = \int_0^{\pi/2} \frac{a^4\sin^4 \theta}{a\cos \theta} (a\cos \theta d \theta) \)
\( I = a^4 \int_0^{\pi/2} \sin^4 \theta d \theta \)
Use the identity \( \sin^2 \theta = \frac{1-\cos 2\theta}{2} \):
\( I = a^4 \int_0^{\pi/2} \left( \frac{1-\cos 2\theta}{2} \right)^2 d \theta \)
\( I = a^4 \int_0^{\pi/2} \frac{1-2\cos 2\theta+\cos^2 2\theta}{4} d \theta \)
Use the identity \( \cos^2 A = \frac{1+\cos 2A}{2} \): So, \( \cos^2 2\theta = \frac{1+\cos 4\theta}{2} \).
\( I = \frac{a^4}{4} \int_0^{\pi/2} \left( 1-2\cos 2\theta + \frac{1+\cos 4\theta}{2} \right) d \theta \)
\( I = \frac{a^4}{8} \int_0^{\pi/2} (2-4\cos 2\theta + 1+\cos 4\theta) d \theta \)
\( I = \frac{a^4}{8} \int_0^{\pi/2} (3-4\cos 2\theta + \cos 4\theta) d \theta \)
Integrate term by term:
\( I = \frac{a^4}{8} \left[ 3\theta - \frac{4\sin 2\theta}{2} + \frac{\sin 4\theta}{4} \right]_0^{\pi/2} \)
\( I = \frac{a^4}{8} \left[ 3\theta - 2\sin 2\theta + \frac{1}{4}\sin 4\theta \right]_0^{\pi/2} \)
Apply the limits:
\( I = \frac{a^4}{8} \left[ \left( 3\frac{\pi}{2} - 2\sin (2\frac{\pi}{2}) + \frac{1}{4}\sin (4\frac{\pi}{2}) \right) - \left( 3(0) - 2\sin (0) + \frac{1}{4}\sin (0) \right) \right] \)
\( I = \frac{a^4}{8} \left[ \left( \frac{3\pi}{2} - 2\sin \pi + \frac{1}{4}\sin 2\pi \right) - (0 - 0 + 0) \right] \)
Since \( \sin \pi = 0 \) and \( \sin 2\pi = 0 \):
\( I = \frac{a^4}{8} \left[ \frac{3\pi}{2} - 0 + 0 \right] = \frac{a^4}{8} \cdot \frac{3\pi}{2} = \frac{3\pi a^4}{16} \)
(iv) Let \( I = \int_8^{15} \frac{d x}{(x-3) \sqrt{x+1}} \)
Use substitution. Let \( t = \sqrt{x+1} \).
Then \( t^2 = x+1 \implies x = t^2-1 \).
Differentiate \( x \) with respect to \( t \): \( dx = 2t dt \).
The term \( (x-3) \) becomes \( (t^2-1-3) = t^2-4 \).
Change the limits. When \( x = 8 \), \( t = \sqrt{8+1} = \sqrt{9} = 3 \).
When \( x = 15 \), \( t = \sqrt{15+1} = \sqrt{16} = 4 \).
Substitute into the integral:
\( I = \int_3^4 \frac{2t dt}{(t^2-4) t} \)
Cancel \( t \) from numerator and denominator:
\( I = \int_3^4 \frac{2 dt}{t^2-4} \)
This is a standard integral form: \( \int \frac{1}{x^2-a^2} dx = \frac{1}{2a} \log \left| \frac{x-a}{x+a} \right| \). Here \( a=2 \).
\( I = 2 \left[ \frac{1}{2(2)} \log \left| \frac{t-2}{t+2} \right| \right]_3^4 \)
\( I = \frac{1}{2} \left[ \log \left| \frac{t-2}{t+2} \right| \right]_3^4 \)
Apply the limits:
\( I = \frac{1}{2} \left( \log \left| \frac{4-2}{4+2} \right| - \log \left| \frac{3-2}{3+2} \right| \right) \)
\( I = \frac{1}{2} \left( \log \left| \frac{2}{6} \right| - \log \left| \frac{1}{5} \right| \right) \)
\( I = \frac{1}{2} \left( \log \left( \frac{1}{3} \right) - \log \left( \frac{1}{5} \right) \right) \)
Using logarithm properties \( \log a - \log b = \log (a/b) \):
\( I = \frac{1}{2} \log \left( \frac{1/3}{1/5} \right) = \frac{1}{2} \log \left( \frac{1}{3} \times 5 \right) = \frac{1}{2} \log \left( \frac{5}{3} \right) \)
(v) Let \( I = \int_{1 / 3}^1 \frac{\left(x-x^3\right)^{1 / 3}}{x^4} d x \)
Factor \( x^3 \) from inside the parenthesis: \( x-x^3 = x^3(\frac{1}{x^2}-1) \).
So, \( (x-x^3)^{1/3} = (x^3(\frac{1}{x^2}-1))^{1/3} = (x^3)^{1/3} (\frac{1}{x^2}-1)^{1/3} = x (\frac{1}{x^2}-1)^{1/3} \).
Substitute this back into the integral:
\( I = \int_{1 / 3}^1 \frac{x (\frac{1}{x^2}-1)^{1/3}}{x^4} d x = \int_{1 / 3}^1 \frac{(\frac{1}{x^2}-1)^{1/3}}{x^3} d x \)
Use substitution. Let \( t = \frac{1}{x^2}-1 \).
Differentiate \( t \) with respect to \( x \): \( dt = -2x^{-3} dx = -\frac{2}{x^3} dx \).
So, \( \frac{1}{x^3} dx = -\frac{1}{2} dt \).
Change the limits. When \( x = \frac{1}{3} \), \( t = \frac{1}{(1/3)^2}-1 = \frac{1}{1/9}-1 = 9-1 = 8 \).
When \( x = 1 \), \( t = \frac{1}{1^2}-1 = 1-1 = 0 \).
Substitute into the integral:
\( I = \int_8^0 t^{1/3} \left( -\frac{1}{2} dt \right) \)
\( I = -\frac{1}{2} \int_8^0 t^{1/3} dt \)
Swap the limits and change the sign:
\( I = \frac{1}{2} \int_0^8 t^{1/3} dt \)
Integrate \( t^{1/3} \): \( \int t^{1/3} dt = \frac{t^{1/3+1}}{1/3+1} = \frac{t^{4/3}}{4/3} = \frac{3}{4} t^{4/3} \).
\( I = \frac{1}{2} \left[ \frac{3}{4} t^{4/3} \right]_0^8 \)
\( I = \frac{3}{8} [t^{4/3}]_0^8 \)
Apply the limits:
\( I = \frac{3}{8} (8^{4/3} - 0^{4/3}) \)
\( I = \frac{3}{8} ((2^3)^{4/3} - 0) = \frac{3}{8} (2^4) = \frac{3}{8} \times 16 = 3 \times 2 = 6 \)
In simple words: These integrals often involve square roots or fractional powers, so the main idea is to pick the right substitution. For expressions like \( \sqrt{a^2-x^2} \), a substitution like \( x=a\sin\theta \) usually works well. For other square root forms, setting the entire root equal to \( t \) or just the term under it can simplify things. Always remember to adjust the integration limits when you substitute.

🎯 Exam Tip: When dealing with integrals involving \( \sqrt{a^2-x^2} \), try trigonometric substitutions like \( x = a\sin\theta \) or \( x = a\cos\theta \). For rational functions involving \( \sqrt{x+k} \), the substitution \( t=\sqrt{x+k} \) is often effective, as it eliminates the radical term.

Question 6. Evaluate the following definite integrals:
(i) \( \int_0^1 \cos ^{-1} x d x \)
(ii) \( \int_0^1 \tan ^{-1} x d x \)
(iii) \( \int_0^1 x \sin ^{-1} x d x \)
(iv) \( \int_0^1 \sin ^{-1}\left(\frac{2 x}{1+x^2}\right) d x \)
Answer:
(i) Let \( I = \int_0^1 \cos ^{-1} x d x \)
We use integration by parts, which is \( \int u dv = uv - \int v du \). Treat \( \cos^{-1} x \) as \( u \) and \( 1 \) as \( dv \).
Let \( u = \cos^{-1} x \implies du = -\frac{1}{\sqrt{1-x^2}} dx \).
Let \( dv = 1 dx \implies v = x \).
\( I = [x \cos^{-1} x]_0^1 - \int_0^1 x \left( -\frac{1}{\sqrt{1-x^2}} \right) dx \)
\( I = (1 \cdot \cos^{-1} 1 - 0 \cdot \cos^{-1} 0) + \int_0^1 \frac{x}{\sqrt{1-x^2}} dx \)
We know \( \cos^{-1} 1 = 0 \) and \( \cos^{-1} 0 = \frac{\pi}{2} \).
\( I = (1 \cdot 0 - 0 \cdot \frac{\pi}{2}) + \int_0^1 x (1-x^2)^{-1/2} dx \)
\( I = 0 + \int_0^1 x (1-x^2)^{-1/2} dx \)
For the remaining integral, use substitution. Let \( t = 1-x^2 \).
Then \( dt = -2x dx \implies x dx = -\frac{1}{2} dt \).
Change the limits. When \( x = 0 \), \( t = 1-0^2 = 1 \).
When \( x = 1 \), \( t = 1-1^2 = 0 \).
Substitute into the integral:
\( I = \int_1^0 t^{-1/2} \left( -\frac{1}{2} dt \right) \)
\( I = -\frac{1}{2} \int_1^0 t^{-1/2} dt \)
Swap the limits and change the sign:
\( I = \frac{1}{2} \int_0^1 t^{-1/2} dt \)
Integrate \( t^{-1/2} \): \( \int t^{-1/2} dt = \frac{t^{1/2}}{1/2} = 2\sqrt{t} \).
\( I = \frac{1}{2} [2\sqrt{t}]_0^1 = [\sqrt{t}]_0^1 \)
Apply the limits:
\( I = \sqrt{1} - \sqrt{0} = 1 - 0 = 1 \)
(ii) Let \( I = \int_0^1 \tan ^{-1} x d x \)
Use integration by parts: \( \int u dv = uv - \int v du \). Treat \( \tan^{-1} x \) as \( u \) and \( 1 \) as \( dv \).
Let \( u = \tan^{-1} x \implies du = \frac{1}{1+x^2} dx \).
Let \( dv = 1 dx \implies v = x \).
\( I = [x \tan^{-1} x]_0^1 - \int_0^1 x \left( \frac{1}{1+x^2} \right) dx \)
\( I = (1 \cdot \tan^{-1} 1 - 0 \cdot \tan^{-1} 0) - \int_0^1 \frac{x}{1+x^2} dx \)
We know \( \tan^{-1} 1 = \frac{\pi}{4} \) and \( \tan^{-1} 0 = 0 \).
\( I = (\frac{\pi}{4} - 0) - \int_0^1 \frac{x}{1+x^2} dx \)
For the remaining integral, use substitution. Let \( t = 1+x^2 \).
Then \( dt = 2x dx \implies x dx = \frac{1}{2} dt \).
Change the limits. When \( x = 0 \), \( t = 1+0^2 = 1 \).
When \( x = 1 \), \( t = 1+1^2 = 2 \).
Substitute into the integral:
\( I = \frac{\pi}{4} - \int_1^2 \frac{1}{t} \left( \frac{1}{2} dt \right) \)
\( I = \frac{\pi}{4} - \frac{1}{2} \int_1^2 \frac{1}{t} dt \)
\( I = \frac{\pi}{4} - \frac{1}{2} [\log |t|]_1^2 \)
Apply the limits:
\( I = \frac{\pi}{4} - \frac{1}{2} (\log 2 - \log 1) \)
Since \( \log 1 = 0 \):
\( I = \frac{\pi}{4} - \frac{1}{2} \log 2 \)
(iii) Let \( I = \int_0^1 x \sin ^{-1} x d x \)
Use integration by parts: \( \int u dv = uv - \int v du \). Treat \( \sin^{-1} x \) as \( u \) and \( x \) as \( dv \).
Let \( u = \sin^{-1} x \implies du = \frac{1}{\sqrt{1-x^2}} dx \).
Let \( dv = x dx \implies v = \frac{x^2}{2} \).
\( I = \left[ \frac{x^2}{2} \sin^{-1} x \right]_0^1 - \int_0^1 \frac{x^2}{2} \left( \frac{1}{\sqrt{1-x^2}} \right) dx \)
\( I = \left( \frac{1^2}{2} \sin^{-1} 1 - \frac{0^2}{2} \sin^{-1} 0 \right) - \frac{1}{2} \int_0^1 \frac{x^2}{\sqrt{1-x^2}} dx \)
We know \( \sin^{-1} 1 = \frac{\pi}{2} \) and \( \sin^{-1} 0 = 0 \).
\( I = \left( \frac{1}{2} \cdot \frac{\pi}{2} - 0 \right) - \frac{1}{2} \int_0^1 \frac{x^2}{\sqrt{1-x^2}} dx \)
\( I = \frac{\pi}{4} - \frac{1}{2} \int_0^1 \frac{x^2}{\sqrt{1-x^2}} dx \)
For the remaining integral, rewrite the numerator: \( x^2 = -(1-x^2)+1 \).
\( \int_0^1 \frac{-(1-x^2)+1}{\sqrt{1-x^2}} dx = \int_0^1 \left( -\sqrt{1-x^2} + \frac{1}{\sqrt{1-x^2}} \right) dx \)
We know \( \int \sqrt{a^2-x^2} dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a} \). So \( \int \sqrt{1-x^2} dx = \frac{x}{2}\sqrt{1-x^2} + \frac{1}{2}\sin^{-1} x \).
And \( \int \frac{1}{\sqrt{1-x^2}} dx = \sin^{-1} x \).
So, \( \int_0^1 \left( -\sqrt{1-x^2} + \frac{1}{\sqrt{1-x^2}} \right) dx = \left[ -(\frac{x}{2}\sqrt{1-x^2} + \frac{1}{2}\sin^{-1} x) + \sin^{-1} x \right]_0^1 \)
\( = \left[ -\frac{x}{2}\sqrt{1-x^2} + \frac{1}{2}\sin^{-1} x \right]_0^1 \)
Apply the limits:
\( = \left( -\frac{1}{2}\sqrt{1-1^2} + \frac{1}{2}\sin^{-1} 1 \right) - \left( -\frac{0}{2}\sqrt{1-0^2} + \frac{1}{2}\sin^{-1} 0 \right) \)
\( = \left( -\frac{1}{2}\sqrt{0} + \frac{1}{2}\cdot\frac{\pi}{2} \right) - (0 + 0) = \frac{\pi}{4} \).
Substitute this back into the main integral:
\( I = \frac{\pi}{4} - \frac{1}{2} \left( \frac{\pi}{4} \right) = \frac{\pi}{4} - \frac{\pi}{8} = \frac{\pi}{8} \)
(iv) Let \( I = \int_0^1 \sin ^{-1}\left(\frac{2 x}{1+x^2}\right) d x \)
Use the trigonometric identity \( \sin^{-1}\left(\frac{2 x}{1+x^2}\right) = 2 \tan^{-1} x \).
So, \( I = \int_0^1 2 \tan^{-1} x d x = 2 \int_0^1 \tan^{-1} x d x \).
From part (ii) of this question, we already calculated \( \int_0^1 \tan^{-1} x d x = \frac{\pi}{4} - \frac{1}{2} \log 2 \).
Substitute this value:
\( I = 2 \left( \frac{\pi}{4} - \frac{1}{2} \log 2 \right) \)
\( I = \frac{\pi}{2} - \log 2 \)
In simple words: For integrals with inverse trigonometric functions, a good strategy is to use integration by parts. For complex expressions within inverse functions, try to use trigonometric identities to simplify them first. For example, \( \sin^{-1}(\frac{2x}{1+x^2}) \) is actually the same as \( 2\tan^{-1}x \), which makes the integral much easier.

🎯 Exam Tip: When faced with inverse trigonometric functions in an integral, always consider integration by parts. Also, look for identities like \( 2\tan^{-1} x = \sin^{-1} \frac{2x}{1+x^2} \) or \( 2\tan^{-1} x = \cos^{-1} \frac{1-x^2}{1+x^2} \) that can simplify the integrand significantly.

Question 7. Evaluate : \( \int_0^\pi e^{2 x} \sin \left(\frac{\pi}{4}+x\right) d x \)
Answer:
Let \( I = \int_0^\pi e^{2 x} \sin \left(\frac{\pi}{4}+x\right) d x \)
We can use the direct formula for integrals of the form \( \int e^{ax} \sin(bx+c) dx = \frac{e^{ax}}{a^2+b^2} [a \sin(bx+c) - b \cos(bx+c)] \).
Here, \( a=2 \), \( b=1 \), and \( c=\frac{\pi}{4} \). So \( a^2+b^2 = 2^2+1^2 = 5 \).
\( I = \left[ \frac{e^{2x}}{5} \left( 2 \sin\left(\frac{\pi}{4}+x\right) - 1 \cos\left(\frac{\pi}{4}+x\right) \right) \right]_0^\pi \)
Apply the limits of integration:
\( I = \frac{1}{5} \left[ e^{2\pi} \left( 2 \sin\left(\frac{\pi}{4}+\pi\right) - \cos\left(\frac{\pi}{4}+\pi\right) \right) - e^{2(0)} \left( 2 \sin\left(\frac{\pi}{4}+0\right) - \cos\left(\frac{\pi}{4}+0\right) \right) \right] \)
Simplify the trigonometric terms:
\( \sin\left(\frac{\pi}{4}+\pi\right) = \sin\left(\frac{5\pi}{4}\right) = -\frac{1}{\sqrt{2}} \)
\( \cos\left(\frac{\pi}{4}+\pi\right) = \cos\left(\frac{5\pi}{4}\right) = -\frac{1}{\sqrt{2}} \)
\( \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \)
\( \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \)
Substitute these values:
\( I = \frac{1}{5} \left[ e^{2\pi} \left( 2 \left(-\frac{1}{\sqrt{2}}\right) - \left(-\frac{1}{\sqrt{2}}\right) \right) - e^0 \left( 2 \left(\frac{1}{\sqrt{2}}\right) - \frac{1}{\sqrt{2}} \right) \right] \)
\( I = \frac{1}{5} \left[ e^{2\pi} \left( -\frac{2}{\sqrt{2}} + \frac{1}{\sqrt{2}} \right) - 1 \left( \frac{2}{\sqrt{2}} - \frac{1}{\sqrt{2}} \right) \right] \)
\( I = \frac{1}{5} \left[ e^{2\pi} \left( -\frac{1}{\sqrt{2}} \right) - 1 \left( \frac{1}{\sqrt{2}} \right) \right] \)
\( I = \frac{1}{5\sqrt{2}} (-e^{2\pi} - 1) = -\frac{e^{2\pi}+1}{5\sqrt{2}} \)
In simple words: When you have an integral with \( e^{ax} \) multiplied by \( \sin(bx) \) or \( \cos(bx) \), there's a special formula you can use. You just plug in the numbers \( a \) and \( b \) from the problem into the formula. After applying the formula, you put in the upper and lower limits to find the final numerical answer. This method saves a lot of time compared to doing integration by parts twice.

🎯 Exam Tip: Memorize the standard formula for integrals of the form \( \int e^{ax} \sin(bx) dx \) and \( \int e^{ax} \cos(bx) dx \). This will help you quickly solve such problems without repeating lengthy integration by parts steps, especially in timed exams.

Question 8. Prove that:
(i) \( \int_0^a \sin ^{-1} \sqrt{\left(\frac{x}{a+x}\right)} d x=a\left(\frac{\pi}{2}-1\right) \)
(ii) \( \int_0^{\pi / 4}(\sqrt{\tan x}+\sqrt{\cot x}) d x=\sqrt{2} \cdot \frac{\pi}{2} \)
Answer:
(i) Let \( I = \int_0^a \sin ^{-1} \sqrt{\frac{x}{a+x}} d x \)
Use substitution. Let \( x = a \tan^2 \theta \).
Differentiate \( x \) with respect to \( \theta \): \( dx = 2a \tan \theta \sec^2 \theta d \theta \).
Change the limits. When \( x = 0 \), \( a \tan^2 \theta = 0 \implies \tan \theta = 0 \implies \theta = 0 \).
When \( x = a \), \( a \tan^2 \theta = a \implies \tan^2 \theta = 1 \implies \theta = \frac{\pi}{4} \).
Now, simplify the term inside the inverse sine function:
\( \sqrt{\frac{x}{a+x}} = \sqrt{\frac{a \tan^2 \theta}{a+a \tan^2 \theta}} = \sqrt{\frac{a \tan^2 \theta}{a(1+\tan^2 \theta)}} = \sqrt{\frac{\tan^2 \theta}{\sec^2 \theta}} = \sqrt{\sin^2 \theta} = \sin \theta \).
So, \( \sin^{-1} \sqrt{\frac{x}{a+x}} = \sin^{-1} (\sin \theta) = \theta \).
Substitute these into the integral:
\( I = \int_0^{\pi/4} \theta (2a \tan \theta \sec^2 \theta) d \theta \)
\( I = 2a \int_0^{\pi/4} \theta \tan \theta \sec^2 \theta d \theta \)
Use integration by parts: \( \int u dv = uv - \int v du \). Let \( u = \theta \) and \( dv = \tan \theta \sec^2 \theta d \theta \).
Then \( du = d \theta \) and \( v = \int \tan \theta \sec^2 \theta d \theta = \frac{\tan^2 \theta}{2} \) (by substitution, let \( w=\tan\theta \)).
\( I = 2a \left[ \left[ \theta \frac{\tan^2 \theta}{2} \right]_0^{\pi/4} - \int_0^{\pi/4} \frac{\tan^2 \theta}{2} d \theta \right] \)
Apply limits to the first term:
\( \left( \frac{\pi}{4} \cdot \frac{\tan^2 (\pi/4)}{2} - 0 \cdot \frac{\tan^2 0}{2} \right) = \frac{\pi}{4} \cdot \frac{1^2}{2} - 0 = \frac{\pi}{8} \).
For the integral term, use \( \tan^2 \theta = \sec^2 \theta - 1 \):
\( \int_0^{\pi/4} \frac{\tan^2 \theta}{2} d \theta = \frac{1}{2} \int_0^{\pi/4} (\sec^2 \theta - 1) d \theta \)
\( = \frac{1}{2} [\tan \theta - \theta]_0^{\pi/4} \)
\( = \frac{1}{2} \left[ (\tan \frac{\pi}{4} - \frac{\pi}{4}) - (\tan 0 - 0) \right] \)
\( = \frac{1}{2} \left[ (1 - \frac{\pi}{4}) - (0 - 0) \right] = \frac{1}{2} - \frac{\pi}{8} \).
Substitute these back into the expression for \( I \):
\( I = 2a \left[ \frac{\pi}{8} - \left( \frac{1}{2} - \frac{\pi}{8} \right) \right] \)
\( I = 2a \left[ \frac{\pi}{8} - \frac{1}{2} + \frac{\pi}{8} \right] \)
\( I = 2a \left[ \frac{2\pi}{8} - \frac{1}{2} \right] = 2a \left[ \frac{\pi}{4} - \frac{1}{2} \right] \)
\( I = a \left( \frac{\pi}{2} - 1 \right) \). This proves the statement.
(ii) Let \( I = \int_0^{\pi / 4}(\sqrt{\tan x}+\sqrt{\cot x}) d x \)
Rewrite \( \cot x \) as \( \frac{1}{\tan x} \):
\( I = \int_0^{\pi / 4} \left( \sqrt{\tan x} + \frac{1}{\sqrt{\tan x}} \right) d x \)
Combine the terms by finding a common denominator:
\( I = \int_0^{\pi / 4} \frac{\tan x+1}{\sqrt{\tan x}} d x \)
Use substitution. Let \( t = \sqrt{\tan x} \).
Then \( t^2 = \tan x \).
Differentiate \( t^2 \) with respect to \( x \): \( 2t dt = \sec^2 x dx = (1+\tan^2 x) dx = (1+(t^2)^2) dx = (1+t^4) dx \).
So, \( dx = \frac{2t dt}{1+t^4} \).
Change the limits. When \( x = 0 \), \( t = \sqrt{\tan 0} = 0 \).
When \( x = \frac{\pi}{4} \), \( t = \sqrt{\tan \frac{\pi}{4}} = \sqrt{1} = 1 \).
Substitute into the integral:
\( I = \int_0^1 \frac{t^2+1}{t} \left( \frac{2t dt}{1+t^4} \right) \)
\( I = 2 \int_0^1 \frac{t^2+1}{t^4+1} dt \)
Divide the numerator and denominator by \( t^2 \):
\( I = 2 \int_0^1 \frac{\frac{t^2+1}{t^2}}{\frac{t^4+1}{t^2}} dt = 2 \int_0^1 \frac{1+\frac{1}{t^2}}{t^2+\frac{1}{t^2}} dt \)
Rewrite the denominator: \( t^2+\frac{1}{t^2} = (t-\frac{1}{t})^2 + 2 \).
\( I = 2 \int_0^1 \frac{1+\frac{1}{t^2}}{\left(t-\frac{1}{t}\right)^2 + 2} dt \)
Use another substitution. Let \( u = t-\frac{1}{t} \).
Differentiate \( u \) with respect to \( t \): \( du = (1+\frac{1}{t^2}) dt \).
Change the limits for \( u \):
When \( t \to 0^+ \), \( u = 0 - \infty = -\infty \).
When \( t = 1 \), \( u = 1 - \frac{1}{1} = 0 \).
Substitute into the integral:
\( I = 2 \int_{-\infty}^0 \frac{du}{u^2 + (\sqrt{2})^2} \)
This is a standard integral form: \( \int \frac{1}{x^2+a^2} dx = \frac{1}{a} \tan^{-1} \frac{x}{a} \). Here \( a = \sqrt{2} \).
\( I = 2 \left[ \frac{1}{\sqrt{2}} \tan^{-1} \frac{u}{\sqrt{2}} \right]_{-\infty}^0 \)
\( I = \frac{2}{\sqrt{2}} \left( \tan^{-1} \frac{0}{\sqrt{2}} - \tan^{-1} \frac{-\infty}{\sqrt{2}} \right) \)
\( I = \sqrt{2} (\tan^{-1} 0 - \tan^{-1} (-\infty)) \)
\( I = \sqrt{2} (0 - (-\frac{\pi}{2})) = \sqrt{2} \cdot \frac{\pi}{2} \). This proves the statement.
In simple words: These problems ask us to prove that an integral equals a specific value. We achieve this by carefully choosing a substitution that simplifies the expression inside the integral. Then, we apply integration rules and substitute the new limits to calculate the integral. For expressions with \( \tan x \) and \( \cot x \), converting them to common terms often helps.

🎯 Exam Tip: For integrals involving \( \sqrt{\tan x} \) or \( \sqrt{\cot x} \), the substitution \( t = \sqrt{\tan x} \) or \( t = \sqrt{\cot x} \) is a very effective strategy. Remember the technique of dividing numerator and denominator by \( t^2 \) after the first substitution to make the denominator suitable for completing the square.

Question 5.
(i) \( \int_0^{\sqrt{2}} \sqrt{2-x^2} d x \)
(ii) \( \int_0^3 \sqrt{9-x^2} d x \)
(iii) \( \int_0^a \frac{x^4}{\sqrt{a^2-x^2}} d x \)
(iv) \( \int_8^{15} \frac{d x}{(x-3) \sqrt{x+1}} \)
(v) \( \int_{1 / 3}^1 \frac{\left(x-x^3\right)^{1 / 3}}{x^4} d x \)
Answer:
(i) We need to evaluate \( \int_0^{\sqrt{2}} \sqrt{2-x^2} d x \).
Let \( x = \sqrt{2} \sin \theta \). Then \( dx = \sqrt{2} \cos \theta d\theta \).
When \( x = 0 \), \( \sqrt{2} \sin \theta = 0 \implies \theta = 0 \).
When \( x = \sqrt{2} \), \( \sqrt{2} \sin \theta = \sqrt{2} \implies \sin \theta = 1 \implies \theta = \frac{\pi}{2} \).
Substitute these into the integral:
\( I = \int_0^{\pi/2} \sqrt{2 - (\sqrt{2}\sin\theta)^2} (\sqrt{2}\cos\theta) d\theta \)
\( I = \int_0^{\pi/2} \sqrt{2 - 2\sin^2\theta} (\sqrt{2}\cos\theta) d\theta \)
\( I = \int_0^{\pi/2} \sqrt{2(1 - \sin^2\theta)} (\sqrt{2}\cos\theta) d\theta \)
\( I = \int_0^{\pi/2} \sqrt{2\cos^2\theta} (\sqrt{2}\cos\theta) d\theta \)
\( I = \int_0^{\pi/2} (\sqrt{2}\cos\theta) (\sqrt{2}\cos\theta) d\theta \) (Since \( 0 \le \theta \le \frac{\pi}{2} \), \( \cos\theta \ge 0 \))
\( I = \int_0^{\pi/2} 2\cos^2\theta d\theta \)
\( I = \int_0^{\pi/2} 2 \left( \frac{1+\cos 2\theta}{2} \right) d\theta \)
\( I = \int_0^{\pi/2} (1+\cos 2\theta) d\theta \)
Now, integrate term by term:
\( I = \left[ \theta + \frac{\sin 2\theta}{2} \right]_0^{\pi/2} \)
\( I = \left( \frac{\pi}{2} + \frac{\sin (2 \cdot \frac{\pi}{2})}{2} \right) - \left( 0 + \frac{\sin (2 \cdot 0)}{2} \right) \)
\( I = \left( \frac{\pi}{2} + \frac{\sin \pi}{2} \right) - (0 + \frac{\sin 0}{2} ) \)
\( I = \left( \frac{\pi}{2} + 0 \right) - (0 + 0) \)
\( I = \frac{\pi}{2} \).

(ii) We need to evaluate \( \int_0^3 \sqrt{9-x^2} d x \).
This integral is of the form \( \int \sqrt{a^2-x^2} dx \), where \( a=3 \).
The formula for this integral is \( \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) \).
So, \( \int_0^3 \sqrt{3^2-x^2} d x = \left[ \frac{x}{2}\sqrt{9-x^2} + \frac{9}{2}\sin^{-1}\left(\frac{x}{3}\right) \right]_0^3 \).
Now, substitute the upper and lower limits:
For the upper limit \( x=3 \):
\( \frac{3}{2}\sqrt{9-3^2} + \frac{9}{2}\sin^{-1}\left(\frac{3}{3}\right) = \frac{3}{2}\sqrt{0} + \frac{9}{2}\sin^{-1}(1) = 0 + \frac{9}{2} \cdot \frac{\pi}{2} = \frac{9\pi}{4} \).
For the lower limit \( x=0 \):
\( \frac{0}{2}\sqrt{9-0^2} + \frac{9}{2}\sin^{-1}\left(\frac{0}{3}\right) = 0 + \frac{9}{2}\sin^{-1}(0) = 0 + 0 = 0 \).
Therefore, the value of the integral is \( \frac{9\pi}{4} - 0 = \frac{9\pi}{4} \).

(iii) We need to evaluate \( \int_0^a \frac{x^4}{\sqrt{a^2-x^2}} d x \).
Let \( x = a \sin \theta \). Then \( dx = a \cos \theta d\theta \).
When \( x = 0 \), \( a \sin \theta = 0 \implies \theta = 0 \).
When \( x = a \), \( a \sin \theta = a \implies \sin \theta = 1 \implies \theta = \frac{\pi}{2} \).
Substitute these into the integral:
\( I = \int_0^{\pi/2} \frac{(a \sin\theta)^4}{\sqrt{a^2-(a \sin\theta)^2}} (a \cos\theta) d\theta \)
\( I = \int_0^{\pi/2} \frac{a^4 \sin^4\theta}{\sqrt{a^2(1-\sin^2\theta)}} (a \cos\theta) d\theta \)
\( I = \int_0^{\pi/2} \frac{a^4 \sin^4\theta}{\sqrt{a^2\cos^2\theta}} (a \cos\theta) d\theta \)
\( I = \int_0^{\pi/2} \frac{a^4 \sin^4\theta}{a \cos\theta} (a \cos\theta) d\theta \) (Since \( 0 \le \theta \le \frac{\pi}{2} \), \( \cos\theta \ge 0 \))
\( I = \int_0^{\pi/2} a^4 \sin^4\theta d\theta \)
\( I = a^4 \int_0^{\pi/2} \sin^4\theta d\theta \)
We use the identity \( \sin^2\theta = \frac{1-\cos 2\theta}{2} \):
\( I = a^4 \int_0^{\pi/2} \left( \frac{1-\cos 2\theta}{2} \right)^2 d\theta \)
\( I = a^4 \int_0^{\pi/2} \frac{1-2\cos 2\theta+\cos^2 2\theta}{4} d\theta \)
We use another identity \( \cos^2 A = \frac{1+\cos 2A}{2} \):
\( I = \frac{a^4}{4} \int_0^{\pi/2} \left( 1-2\cos 2\theta + \frac{1+\cos 4\theta}{2} \right) d\theta \)
\( I = \frac{a^4}{4} \int_0^{\pi/2} \left( \frac{2-4\cos 2\theta + 1+\cos 4\theta}{2} \right) d\theta \)
\( I = \frac{a^4}{8} \int_0^{\pi/2} (3-4\cos 2\theta+\cos 4\theta) d\theta \)
Now, integrate term by term:
\( I = \frac{a^4}{8} \left[ 3\theta - \frac{4\sin 2\theta}{2} + \frac{\sin 4\theta}{4} \right]_0^{\pi/2} \)
\( I = \frac{a^4}{8} \left[ 3\theta - 2\sin 2\theta + \frac{\sin 4\theta}{4} \right]_0^{\pi/2} \)
For the upper limit \( \theta=\frac{\pi}{2} \):
\( 3\left(\frac{\pi}{2}\right) - 2\sin\left(2\frac{\pi}{2}\right) + \frac{\sin\left(4\frac{\pi}{2}\right)}{4} = \frac{3\pi}{2} - 2\sin\pi + \frac{\sin 2\pi}{4} = \frac{3\pi}{2} - 0 + 0 = \frac{3\pi}{2} \).
For the lower limit \( \theta=0 \):
\( 3(0) - 2\sin(0) + \frac{\sin(0)}{4} = 0 - 0 + 0 = 0 \).
Therefore, \( I = \frac{a^4}{8} \left( \frac{3\pi}{2} - 0 \right) = \frac{3\pi a^4}{16} \).

(iv) We need to evaluate \( \int_8^{15} \frac{d x}{(x-3) \sqrt{x+1}} \).
Let \( t = \sqrt{x+1} \). Then \( t^2 = x+1 \), so \( x = t^2-1 \). Also, \( dx = 2t dt \).
When \( x=8 \), \( t = \sqrt{8+1} = \sqrt{9} = 3 \).
When \( x=15 \), \( t = \sqrt{15+1} = \sqrt{16} = 4 \).
Substitute these into the integral:
\( I = \int_3^4 \frac{2t dt}{((t^2-1)-3)t} \)
\( I = \int_3^4 \frac{2t dt}{(t^2-4)t} \)
\( I = \int_3^4 \frac{2 dt}{t^2-4} \)
\( I = \int_3^4 \frac{2 dt}{t^2-2^2} \).
This is of the form \( \int \frac{1}{x^2-a^2} dx = \frac{1}{2a} \log \left| \frac{x-a}{x+a} \right| \). Here \( x=t \) and \( a=2 \).
\( I = 2 \cdot \frac{1}{2 \cdot 2} \left[ \log \left| \frac{t-2}{t+2} \right| \right]_3^4 \)
\( I = \frac{1}{2} \left[ \log \left| \frac{t-2}{t+2} \right| \right]_3^4 \)
Now, substitute the upper and lower limits:
\( I = \frac{1}{2} \left( \log \left| \frac{4-2}{4+2} \right| - \log \left| \frac{3-2}{3+2} \right| \right) \)
\( I = \frac{1}{2} \left( \log \left| \frac{2}{6} \right| - \log \left| \frac{1}{5} \right| \right) \)
\( I = \frac{1}{2} \left( \log \left( \frac{1}{3} \right) - \log \left( \frac{1}{5} \right) \right) \)
Using the logarithm property \( \log A - \log B = \log \frac{A}{B} \):
\( I = \frac{1}{2} \log \left( \frac{1/3}{1/5} \right) \)
\( I = \frac{1}{2} \log \left( \frac{1}{3} \cdot 5 \right) \)
\( I = \frac{1}{2} \log \left( \frac{5}{3} \right) \).

(v) We need to evaluate \( \int_{1 / 3}^1 \frac{\left(x-x^3\right)^{1 / 3}}{x^4} d x \).
First, rewrite the integrand:
\( I = \int_{1/3}^1 \frac{\left[ x^3\left(\frac{1}{x^2}-1\right) \right]^{1/3}}{x^4} dx \)
\( I = \int_{1/3}^1 \frac{x \left(\frac{1}{x^2}-1\right)^{1/3}}{x^4} dx \)
\( I = \int_{1/3}^1 \frac{\left(\frac{1}{x^2}-1\right)^{1/3}}{x^3} dx \)
Now, let \( t = \frac{1}{x^2}-1 \).
Then \( \frac{dt}{dx} = -2x^{-3} = -\frac{2}{x^3} \).
This means \( \frac{1}{x^3} dx = -\frac{1}{2} dt \).
Change the limits of integration:
When \( x=1 \), \( t = \frac{1}{1^2}-1 = 1-1 = 0 \).
When \( x=\frac{1}{3} \), \( t = \frac{1}{(1/3)^2}-1 = \frac{1}{1/9}-1 = 9-1 = 8 \).
Substitute these into the integral:
\( I = \int_8^0 t^{1/3} \left(-\frac{1}{2}\right) dt \)
\( I = -\frac{1}{2} \int_8^0 t^{1/3} dt \)
To integrate \( t^{1/3} \), we use the power rule \( \int x^n dx = \frac{x^{n+1}}{n+1} \):
\( I = -\frac{1}{2} \left[ \frac{t^{1/3+1}}{1/3+1} \right]_8^0 \)
\( I = -\frac{1}{2} \left[ \frac{t^{4/3}}{4/3} \right]_8^0 \)
\( I = -\frac{1}{2} \cdot \frac{3}{4} \left[ t^{4/3} \right]_8^0 \)
\( I = -\frac{3}{8} \left[ 0^{4/3} - 8^{4/3} \right] \)
\( I = -\frac{3}{8} \left[ 0 - (2^3)^{4/3} \right] \)
\( I = -\frac{3}{8} [ -2^4 ] \)
\( I = -\frac{3}{8} (-16) \)
\( I = 6 \).
In simple words: For definite integrals, first find the indefinite integral, then plug in the upper and lower limits. Remember to change the limits when you use substitution, or switch back to the original variable before applying the limits. It helps to simplify the expression before integrating.

🎯 Exam Tip: Always carefully change the limits of integration when performing a substitution in a definite integral. A common mistake is to forget this step and apply the original limits to the new variable, leading to an incorrect answer.

Question 6.
(i) \( \int_0^1 \cos ^{-1} x d x \)
(ii) \( \int_0^1 \tan ^{-1} x d x \)
(iii) \( \int_0^1 x \sin ^{-1} x d x \)
(iv) \( \int_0^1 \sin ^{-1}\left(\frac{2 x}{1+x^2}\right) d x \)
Answer:
(i) We need to evaluate \( \int_0^1 \cos ^{-1} x d x \).
We use integration by parts, \( \int u dv = uv - \int v du \).
Let \( u = \cos^{-1} x \) and \( dv = 1 dx \).
Then \( du = -\frac{1}{\sqrt{1-x^2}} dx \) and \( v = x \).
So, \( I = \left[ x \cos^{-1} x \right]_0^1 - \int_0^1 x \left( -\frac{1}{\sqrt{1-x^2}} \right) dx \)
\( I = \left[ x \cos^{-1} x \right]_0^1 + \int_0^1 \frac{x}{\sqrt{1-x^2}} dx \).
First, evaluate the definite part:
\( [ x \cos^{-1} x ]_0^1 = (1 \cdot \cos^{-1} 1) - (0 \cdot \cos^{-1} 0) = (1 \cdot 0) - (0) = 0 \).
So, \( I = \int_0^1 \frac{x}{\sqrt{1-x^2}} dx \).
For the remaining integral, let \( t = 1-x^2 \). Then \( dt = -2x dx \), which means \( x dx = -\frac{1}{2} dt \).
Change the limits of integration:
When \( x=0 \), \( t = 1-0^2 = 1 \).
When \( x=1 \), \( t = 1-1^2 = 0 \).
Substitute these into the integral:
\( I = \int_1^0 \frac{1}{\sqrt{t}} \left(-\frac{1}{2}\right) dt \)
\( I = -\frac{1}{2} \int_1^0 t^{-1/2} dt \).
Now integrate \( t^{-1/2} \):
\( I = -\frac{1}{2} \left[ \frac{t^{1/2}}{1/2} \right]_1^0 \)
\( I = -\frac{1}{2} [2\sqrt{t}]_1^0 \)
\( I = -[\sqrt{t}]_1^0 \)
\( I = -(\sqrt{0} - \sqrt{1}) \)
\( I = -(0 - 1) \)
\( I = 1 \).

(ii) We need to evaluate \( \int_0^1 \tan ^{-1} x d x \).
We use integration by parts, \( \int u dv = uv - \int v du \).
Let \( u = \tan^{-1} x \) and \( dv = 1 dx \).
Then \( du = \frac{1}{1+x^2} dx \) and \( v = x \).
So, \( I = \left[ x \tan^{-1} x \right]_0^1 - \int_0^1 x \left( \frac{1}{1+x^2} \right) dx \)
\( I = \left[ x \tan^{-1} x \right]_0^1 - \int_0^1 \frac{x}{1+x^2} dx \).
First, evaluate the definite part:
\( [ x \tan^{-1} x ]_0^1 = (1 \cdot \tan^{-1} 1) - (0 \cdot \tan^{-1} 0) = \frac{\pi}{4} - 0 = \frac{\pi}{4} \).
So, \( I = \frac{\pi}{4} - \int_0^1 \frac{x}{1+x^2} dx \).
For the remaining integral, let \( w = 1+x^2 \). Then \( dw = 2x dx \), which means \( x dx = \frac{1}{2} dw \).
Change the limits of integration:
When \( x=0 \), \( w = 1+0^2 = 1 \).
When \( x=1 \), \( w = 1+1^2 = 2 \).
Substitute these into the integral:
\( I = \frac{\pi}{4} - \int_1^2 \frac{1}{w} \cdot \frac{1}{2} dw \)
\( I = \frac{\pi}{4} - \frac{1}{2} \int_1^2 \frac{1}{w} dw \)
Now integrate \( \frac{1}{w} \):
\( I = \frac{\pi}{4} - \frac{1}{2} [\log |w|]_1^2 \)
\( I = \frac{\pi}{4} - \frac{1}{2} (\log 2 - \log 1) \)
\( I = \frac{\pi}{4} - \frac{1}{2} (\log 2 - 0) \)
\( I = \frac{\pi}{4} - \frac{1}{2} \log 2 \).

(iii) We need to evaluate \( \int_0^1 x \sin ^{-1} x d x \).
We use integration by parts, \( \int u dv = uv - \int v du \).
Let \( u = \sin^{-1} x \) and \( dv = x dx \).
Then \( du = \frac{1}{\sqrt{1-x^2}} dx \) and \( v = \frac{x^2}{2} \).
So, \( I = \left[ \frac{x^2}{2} \sin^{-1} x \right]_0^1 - \int_0^1 \frac{x^2}{2\sqrt{1-x^2}} dx \).
First, evaluate the definite part:
\( \left[ \frac{x^2}{2} \sin^{-1} x \right]_0^1 = \left( \frac{1^2}{2} \sin^{-1} 1 \right) - \left( \frac{0^2}{2} \sin^{-1} 0 \right) = \frac{1}{2} \cdot \frac{\pi}{2} - 0 = \frac{\pi}{4} \).
So, \( I = \frac{\pi}{4} - \frac{1}{2} \int_0^1 \frac{x^2}{\sqrt{1-x^2}} dx \).
For the remaining integral, let \( x = \sin \theta \). Then \( dx = \cos \theta d\theta \).
When \( x=0 \), \( \sin \theta = 0 \implies \theta = 0 \).
When \( x=1 \), \( \sin \theta = 1 \implies \theta = \frac{\pi}{2} \).
The integral becomes \( \int_0^{\pi/2} \frac{\sin^2\theta}{\sqrt{1-\sin^2\theta}} \cos\theta d\theta = \int_0^{\pi/2} \frac{\sin^2\theta}{\cos\theta} \cos\theta d\theta = \int_0^{\pi/2} \sin^2\theta d\theta \).
We use the identity \( \sin^2\theta = \frac{1-\cos 2\theta}{2} \):
\( \int_0^{\pi/2} \frac{1-\cos 2\theta}{2} d\theta = \frac{1}{2} \left[ \theta - \frac{\sin 2\theta}{2} \right]_0^{\pi/2} \).
Substitute the limits:
\( \frac{1}{2} \left( \left( \frac{\pi}{2} - \frac{\sin \pi}{2} \right) - \left( 0 - \frac{\sin 0}{2} \right) \right) \)
\( \frac{1}{2} \left( \left( \frac{\pi}{2} - 0 \right) - (0 - 0) \right) = \frac{1}{2} \cdot \frac{\pi}{2} = \frac{\pi}{4} \).
Now substitute this back into the main integral:
\( I = \frac{\pi}{4} - \frac{1}{2} \left( \frac{\pi}{4} \right) \)
\( I = \frac{\pi}{4} - \frac{\pi}{8} \)
\( I = \frac{2\pi - \pi}{8} = \frac{\pi}{8} \).

(iv) We need to evaluate \( \int_0^1 \sin ^{-1}\left(\frac{2 x}{1+x^2}\right) d x \).
We know the identity \( \sin^{-1}\left(\frac{2x}{1+x^2}\right) = 2\tan^{-1}x \).
So, \( I = \int_0^1 2\tan^{-1}x d x \).
We use integration by parts, \( \int u dv = uv - \int v du \).
Let \( u = 2\tan^{-1}x \) and \( dv = 1 dx \).
Then \( du = \frac{2}{1+x^2} dx \) and \( v = x \).
So, \( I = \left[ 2x \tan^{-1}x \right]_0^1 - \int_0^1 x \left( \frac{2}{1+x^2} \right) dx \)
\( I = \left[ 2x \tan^{-1}x \right]_0^1 - 2\int_0^1 \frac{x}{1+x^2} dx \).
First, evaluate the definite part:
\( [ 2x \tan^{-1}x ]_0^1 = (2 \cdot 1 \cdot \tan^{-1}1) - (2 \cdot 0 \cdot \tan^{-1}0) = 2 \cdot \frac{\pi}{4} - 0 = \frac{\pi}{2} \).
So, \( I = \frac{\pi}{2} - 2\int_0^1 \frac{x}{1+x^2} dx \).
For the remaining integral, let \( w = 1+x^2 \). Then \( dw = 2x dx \), which means \( x dx = \frac{1}{2} dw \).
Change the limits of integration:
When \( x=0 \), \( w = 1+0^2 = 1 \).
When \( x=1 \), \( w = 1+1^2 = 2 \).
Substitute these into the integral:
\( I = \frac{\pi}{2} - 2 \int_1^2 \frac{1}{w} \cdot \frac{1}{2} dw \)
\( I = \frac{\pi}{2} - \int_1^2 \frac{1}{w} dw \)
Now integrate \( \frac{1}{w} \):
\( I = \frac{\pi}{2} - [\log |w|]_1^2 \)
\( I = \frac{\pi}{2} - (\log 2 - \log 1) \)
\( I = \frac{\pi}{2} - (\log 2 - 0) \)
\( I = \frac{\pi}{2} - \log 2 \).
In simple words: When you see inverse trigonometric functions inside an integral, integration by parts is often the best strategy. Remember the common derivative formulas for inverse functions, and sometimes substitution can simplify the remaining integral.

🎯 Exam Tip: Recognizing standard integral forms or identities like \( \sin^{-1}\left(\frac{2x}{1+x^2}\right) = 2\tan^{-1}x \) can save significant time and make complex integrals much simpler to solve.

Question 7. Evaluate : \( \int_0^\pi e^{2 x} \sin \left(\frac{\pi}{4}+x\right) d x \)
Answer: We need to evaluate \( I = \int_0^\pi e^{2 x} \sin \left(\frac{\pi}{4}+x\right) d x \).
We can use the standard formula for integrals of the form \( \int e^{ax} \sin(bx+c) dx = \frac{e^{ax}}{a^2+b^2} [a \sin(bx+c) - b \cos(bx+c)] \).
In our case, \( a=2 \), \( b=1 \), and \( c=\frac{\pi}{4} \).
So, the indefinite integral is:
\( \int e^{2x} \sin\left(\frac{\pi}{4}+x\right) dx = \frac{e^{2x}}{2^2+1^2} \left[ 2 \sin\left(\frac{\pi}{4}+x\right) - 1 \cos\left(\frac{\pi}{4}+x\right) \right] \)
\( = \frac{e^{2x}}{5} \left[ 2 \sin\left(\frac{\pi}{4}+x\right) - \cos\left(\frac{\pi}{4}+x\right) \right] \).
Now, we evaluate this definite integral from \( 0 \) to \( \pi \):
\( I = \frac{1}{5} \left[ e^{2x} \left( 2 \sin\left(\frac{\pi}{4}+x\right) - \cos\left(\frac{\pi}{4}+x\right) \right) \right]_0^\pi \)
\( I = \frac{1}{5} \left( e^{2\pi} \left( 2 \sin\left(\frac{\pi}{4}+\pi\right) - \cos\left(\frac{\pi}{4}+\pi\right) \right) - e^{2 \cdot 0} \left( 2 \sin\left(\frac{\pi}{4}+0\right) - \cos\left(\frac{\pi}{4}+0\right) \right) \right) \)
\( I = \frac{1}{5} \left( e^{2\pi} \left( 2 \sin\left(\frac{5\pi}{4}\right) - \cos\left(\frac{5\pi}{4}\right) \right) - e^0 \left( 2 \sin\left(\frac{\pi}{4}\right) - \cos\left(\frac{\pi}{4}\right) \right) \right) \)
We know \( \sin\left(\frac{5\pi}{4}\right) = -\frac{1}{\sqrt{2}} \) and \( \cos\left(\frac{5\pi}{4}\right) = -\frac{1}{\sqrt{2}} \).
Also, \( \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \) and \( \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \). And \( e^0 = 1 \).
\( I = \frac{1}{5} \left( e^{2\pi} \left( 2 \left(-\frac{1}{\sqrt{2}}\right) - \left(-\frac{1}{\sqrt{2}}\right) \right) - 1 \left( 2 \left(\frac{1}{\sqrt{2}}\right) - \frac{1}{\sqrt{2}} \right) \right) \)
\( I = \frac{1}{5} \left( e^{2\pi} \left( -\frac{2}{\sqrt{2}} + \frac{1}{\sqrt{2}} \right) - \left( \frac{2}{\sqrt{2}} - \frac{1}{\sqrt{2}} \right) \right) \)
\( I = \frac{1}{5} \left( e^{2\pi} \left( -\frac{1}{\sqrt{2}} \right) - \left( \frac{1}{\sqrt{2}} \right) \right) \)
\( I = \frac{1}{5} \left( -\frac{e^{2\pi}}{\sqrt{2}} - \frac{1}{\sqrt{2}} \right) \)
\( I = -\frac{1}{5\sqrt{2}} (e^{2\pi} + 1) \). This type of integral often uses integration by parts twice to solve for I.
In simple words: This integral can be solved using a special formula for exponential and trigonometric functions. You find the general solution and then put in the upper and lower limits to get the final answer. Remember to handle the trigonometric values at different angles carefully.

🎯 Exam Tip: For integrals of the form \( \int e^{ax}\sin(bx) dx \) or \( \int e^{ax}\cos(bx) dx \), memorize the direct formulas or be prepared to use integration by parts twice and solve for the integral. Pay close attention to signs and trigonometric values at the limits.

Question 8. Prove that:
(i) \( \int_0^a \sin ^{-1} \sqrt{\left(\frac{x}{a+x}\right)} d x=a\left(\frac{\pi}{2}-1\right) \)
(ii) \( \int_0^{\pi / 4}(\sqrt{\tan x}+\sqrt{\cot x}) d x=\sqrt{2} \cdot \frac{\pi}{2} \)
Answer:
(i) We need to prove that \( \int_0^a \sin ^{-1} \sqrt{\left(\frac{x}{a+x}\right)} d x=a\left(\frac{\pi}{2}-1\right) \).
Let \( I = \int_0^a \sin ^{-1} \sqrt{\left(\frac{x}{a+x}\right)} d x \).
Let \( x = a \tan^2\theta \). Then \( dx = 2a \tan\theta \sec^2\theta d\theta \).
Change the limits of integration:
When \( x=0 \), \( a \tan^2\theta = 0 \implies \tan\theta = 0 \implies \theta = 0 \).
When \( x=a \), \( a \tan^2\theta = a \implies \tan^2\theta = 1 \implies \tan\theta = 1 \implies \theta = \frac{\pi}{4} \) (since \( \theta \ge 0 \)).
Substitute these into the integral:
\( I = \int_0^{\pi/4} \sin^{-1} \sqrt{\frac{a \tan^2\theta}{a+a \tan^2\theta}} (2a \tan\theta \sec^2\theta d\theta) \)
\( I = \int_0^{\pi/4} \sin^{-1} \sqrt{\frac{a \tan^2\theta}{a(1+\tan^2\theta)}} (2a \tan\theta \sec^2\theta d\theta) \)
\( I = \int_0^{\pi/4} \sin^{-1} \sqrt{\frac{\tan^2\theta}{\sec^2\theta}} (2a \tan\theta \sec^2\theta d\theta) \)
\( I = \int_0^{\pi/4} \sin^{-1} \sqrt{\sin^2\theta} (2a \tan\theta \sec^2\theta d\theta) \)
\( I = \int_0^{\pi/4} \sin^{-1} (\sin\theta) (2a \tan\theta \sec^2\theta d\theta) \) (Since \( 0 \le \theta \le \frac{\pi}{4} \), \( \sin\theta \ge 0 \))
\( I = \int_0^{\pi/4} \theta (2a \tan\theta \sec^2\theta d\theta) \)
\( I = 2a \int_0^{\pi/4} \theta (\tan\theta \sec^2\theta) d\theta \).
Now, we use integration by parts, \( \int u dv = uv - \int v du \).
Let \( u = \theta \) and \( dv = \tan\theta \sec^2\theta d\theta \).
Then \( du = d\theta \). To find \( v \), integrate \( \tan\theta \sec^2\theta \). Let \( p = \tan\theta \), then \( dp = \sec^2\theta d\theta \), so \( \int p dp = \frac{p^2}{2} = \frac{\tan^2\theta}{2} \). Thus, \( v = \frac{\tan^2\theta}{2} \).
\( I = 2a \left( \left[ \theta \frac{\tan^2\theta}{2} \right]_0^{\pi/4} - \int_0^{\pi/4} \frac{\tan^2\theta}{2} d\theta \right) \)
\( I = a \left[ \theta \tan^2\theta \right]_0^{\pi/4} - a \int_0^{\pi/4} \tan^2\theta d\theta \).
First, evaluate the definite part:
\( a \left( \frac{\pi}{4} \tan^2\left(\frac{\pi}{4}\right) - 0 \cdot \tan^2(0) \right) = a \left( \frac{\pi}{4} \cdot 1^2 - 0 \right) = \frac{a\pi}{4} \).
So, \( I = \frac{a\pi}{4} - a \int_0^{\pi/4} \tan^2\theta d\theta \).
Now, for the remaining integral, use the identity \( \tan^2\theta = \sec^2\theta - 1 \):
\( a \int_0^{\pi/4} (\sec^2\theta - 1) d\theta = a \left[ \tan\theta - \theta \right]_0^{\pi/4} \)
\( = a \left( (\tan\frac{\pi}{4} - \frac{\pi}{4}) - (\tan 0 - 0) \right) \)
\( = a \left( (1 - \frac{\pi}{4}) - (0 - 0) \right) = a \left( 1 - \frac{\pi}{4} \right) \).
Substitute this back into the expression for \( I \):
\( I = \frac{a\pi}{4} - a \left( 1 - \frac{\pi}{4} \right) \)
\( I = \frac{a\pi}{4} - a + \frac{a\pi}{4} \)
\( I = \frac{2a\pi}{4} - a \)
\( I = \frac{a\pi}{2} - a \)
\( I = a\left(\frac{\pi}{2} - 1\right) \). This matches the RHS, so the proof is complete.

(ii) We need to prove that \( \int_0^{\pi / 4}(\sqrt{\tan x}+\sqrt{\cot x}) d x=\sqrt{2} \cdot \frac{\pi}{2} \).
Let \( I = \int_0^{\pi / 4}(\sqrt{\tan x}+\sqrt{\cot x}) d x \).
Rewrite \( \sqrt{\cot x} \) as \( \frac{1}{\sqrt{\tan x}} \):
\( I = \int_0^{\pi/4} \left( \sqrt{\tan x} + \frac{1}{\sqrt{\tan x}} \right) d x \)
\( I = \int_0^{\pi/4} \frac{\tan x+1}{\sqrt{\tan x}} d x \).
Let \( t = \sqrt{\tan x} \). Then \( t^2 = \tan x \).
Differentiating both sides with respect to \( x \): \( 2t \frac{dt}{dx} = \sec^2 x \).
So, \( dx = \frac{2t dt}{\sec^2 x} = \frac{2t dt}{1+\tan^2 x} = \frac{2t dt}{1+t^4} \).
Change the limits of integration:
When \( x=0 \), \( t = \sqrt{\tan 0} = 0 \).
When \( x=\frac{\pi}{4} \), \( t = \sqrt{\tan \frac{\pi}{4}} = \sqrt{1} = 1 \).
Substitute these into the integral:
\( I = \int_0^1 \frac{t^2+1}{t} \cdot \frac{2t dt}{1+t^4} \)
\( I = 2 \int_0^1 \frac{t^2+1}{t^4+1} dt \).
To simplify the integrand, divide both the numerator and denominator by \( t^2 \):
\( I = 2 \int_0^1 \frac{\frac{t^2}{t^2}+\frac{1}{t^2}}{\frac{t^4}{t^2}+\frac{1}{t^2}} dt = 2 \int_0^1 \frac{1+\frac{1}{t^2}}{t^2+\frac{1}{t^2}} dt \).
Now, let \( u = t - \frac{1}{t} \). Then \( du = \left( 1+\frac{1}{t^2} \right) dt \).
Also, \( u^2 = \left( t - \frac{1}{t} \right)^2 = t^2 + \frac{1}{t^2} - 2 \). So, \( t^2 + \frac{1}{t^2} = u^2+2 \).
Change the limits for \( u \):
When \( t=0 \), \( u = 0 - \frac{1}{0} \), which approaches \( -\infty \).
When \( t=1 \), \( u = 1 - \frac{1}{1} = 0 \).
Substitute these into the integral:
\( I = 2 \int_{-\infty}^0 \frac{du}{u^2+2} = 2 \int_{-\infty}^0 \frac{du}{u^2+(\sqrt{2})^2} \).
This is a standard integral form \( \int \frac{1}{x^2+a^2} dx = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) \). Here \( a=\sqrt{2} \).
\( I = 2 \cdot \frac{1}{\sqrt{2}} \left[ \tan^{-1}\left(\frac{u}{\sqrt{2}}\right) \right]_{-\infty}^0 \)
\( I = \sqrt{2} \left( \tan^{-1}\left(\frac{0}{\sqrt{2}}\right) - \tan^{-1}\left(\frac{-\infty}{\sqrt{2}}\right) \right) \)
\( I = \sqrt{2} \left( \tan^{-1}(0) - \tan^{-1}(-\infty) \right) \)
\( I = \sqrt{2} \left( 0 - \left(-\frac{\pi}{2}\right) \right) \)
\( I = \sqrt{2} \cdot \frac{\pi}{2} \). This matches the RHS, so the proof is complete.
In simple words: For integrals where you need to prove a result, substitute a variable (like \(x = a \tan^2\theta\)) that simplifies the expression inside the inverse function. For integrals with square roots of tan and cot, convert them to a common base (like tan x) and then use substitution, often dividing by \(t^2\) to create a simpler form for another substitution.

🎯 Exam Tip: Proof-based questions require showing every step clearly. When dealing with \( \sqrt{\tan x} \) and \( \sqrt{\cot x} \), the substitution \( t = \sqrt{\tan x} \) is often very effective. Remember standard identities and integral formulas to simplify your work.