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

S Chand Class 12 ICSE Maths Solutions Chapter 16 Definite Integrals Ex 16(a)

Evaluate the following integrals :

Question 1.
(i) \( \int_{\pi / 4}^{\pi / 2} \cot x d x \)
(ii) \( \int_{\pi / 6}^{\pi / 3} \frac{d x}{\sin 2 x} \)
(iii) \( \int_0^{\pi / 4}\left(2 \sec ^2 x+x^3+2\right) d x \)
(iv) \( \int_0^\pi\left(\sin ^2 \frac{x}{2}-\cos ^2 \frac{x}{2}\right) d x \)
Answer:
(i) We need to find the definite integral of \( \cot x \) from \( \frac{\pi}{4} \) to \( \frac{\pi}{2} \).
The integral of \( \cot x \) is \( \log |\sin x| \).
So, \( \int_{\pi / 4}^{\pi / 2} \cot x d x = [\log |\sin x|]_{\pi / 4}^{\pi / 2} \)
Now, we apply the limits:
\( = \log \sin (\frac{\pi}{2}) - \log \sin (\frac{\pi}{4}) \)
\( = \log 1 - \log (\frac{1}{\sqrt{2}}) \)
\( = 0 - \log (2^{-1/2}) \)
\( = - (-\frac{1}{2}) \log 2 \)
\( = \frac{1}{2} \log 2 \) To remove the negative sign from the exponent, we bring the term to the denominator.
(ii) We need to evaluate \( \int_{\pi / 6}^{\pi / 3} \frac{d x}{\sin 2 x} \).
We know that \( \frac{1}{\sin 2x} = \operatorname{cosec} 2x \).
So, \( \int_{\pi / 6}^{\pi / 3} \operatorname{cosec} 2x d x \).
Let \( t = 2x \), then \( dt = 2 dx \implies dx = \frac{dt}{2} \).
When \( x = \frac{\pi}{6} \), \( t = 2(\frac{\pi}{6}) = \frac{\pi}{3} \).
When \( x = \frac{\pi}{3} \), \( t = 2(\frac{\pi}{3}) = \frac{2\pi}{3} \).
The integral becomes:
\( = \int_{\pi / 3}^{2\pi / 3} \operatorname{cosec} t \frac{dt}{2} \)
\( = \frac{1}{2} [\log |\tan (\frac{t}{2})|]_{\pi / 3}^{2\pi / 3} \)
\( = \frac{1}{2} [\log |\tan (\frac{2\pi}{6})| - \log |\tan (\frac{\pi}{6})|] \)
\( = \frac{1}{2} [\log |\tan (\frac{\pi}{3})| - \log |\tan (\frac{\pi}{6})|] \)
\( = \frac{1}{2} [\log |\sqrt{3}| - \log |\frac{1}{\sqrt{3}}|] \)
\( = \frac{1}{2} [\log \sqrt{3} - \log (3^{-1/2})] \)
\( = \frac{1}{2} [\log \sqrt{3} + \frac{1}{2} \log 3] \)
\( = \frac{1}{2} [\frac{1}{2} \log 3 + \frac{1}{2} \log 3] \)
\( = \frac{1}{2} [\log 3] \)
\( = \frac{1}{2} \log 3 \)
(iii) We need to evaluate \( \int_0^{\pi / 4}(2 \sec ^2 x+x^3+2) d x \).
We can integrate each term separately:
\( = [2 \tan x + \frac{x^4}{4} + 2x]_0^{\pi / 4} \)
Now, we apply the limits:
\( = (2 \tan (\frac{\pi}{4}) + \frac{(\frac{\pi}{4})^4}{4} + 2(\frac{\pi}{4})) - (2 \tan 0 + \frac{0^4}{4} + 2(0)) \)
\( = (2(1) + \frac{\pi^4}{256 \times 4} + \frac{\pi}{2}) - (0 + 0 + 0) \)
\( = 2 + \frac{\pi^4}{1024} + \frac{\pi}{2} \) The tangent of \( \frac{\pi}{4} \) is 1, and tangent of 0 is 0.
(iv) We need to evaluate \( \int_0^\pi\left(\sin ^2 \frac{x}{2}-\cos ^2 \frac{x}{2}\right) d x \).
We know the identity \( \cos 2\theta = \cos^2 \theta - \sin^2 \theta \).
So, \( \sin^2 \frac{x}{2} - \cos^2 \frac{x}{2} = -(\cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}) = -\cos (2 \times \frac{x}{2}) = -\cos x \).
The integral becomes:
\( = \int_0^\pi (-\cos x) d x \)
\( = -[\sin x]_0^\pi \)
\( = -(\sin \pi - \sin 0) \)
\( = -(0 - 0) \)
\( = 0 \)
In simple words: For definite integrals, first find the anti-derivative, then put in the top limit number, then subtract what you get when you put in the bottom limit number. Remember to use trigonometric identities to simplify expressions before integrating.

🎯 Exam Tip: Always simplify the integrand using trigonometric identities before integration, as it often makes the problem much easier to solve. Remember the fundamental theorem of calculus for evaluating definite integrals.

Question 2.
(i) \( \int_0^1 \frac{d x}{2 x-3} \)
(ii) \( \int_1^3 \frac{d x}{7-2 x} \)
Answer:
(i) We need to evaluate \( \int_0^1 \frac{d x}{2 x-3} \).
The integral of \( \frac{1}{ax+b} \) is \( \frac{1}{a} \log |ax+b| \).
So, \( \int_0^1 \frac{d x}{2 x-3} = [\frac{1}{2} \log |2x-3|]_0^1 \)
Now, apply the limits:
\( = \frac{1}{2} [\log |2(1)-3| - \log |2(0)-3|] \)
\( = \frac{1}{2} [\log |-1| - \log |-3|] \)
\( = \frac{1}{2} [\log 1 - \log 3] \)
\( = \frac{1}{2} [0 - \log 3] \)
\( = -\frac{1}{2} \log 3 \) Remember that \( \log 1 = 0 \).
(ii) We need to evaluate \( \int_1^3 \frac{d x}{7-2 x} \).
Using the same integral formula:
\( \int_1^3 \frac{d x}{7-2 x} = [\frac{1}{-2} \log |7-2x|]_1^3 \)
Now, apply the limits:
\( = -\frac{1}{2} [\log |7-2(3)| - \log |7-2(1)|] \)
\( = -\frac{1}{2} [\log |7-6| - \log |7-2|] \)
\( = -\frac{1}{2} [\log |1| - \log |5|] \)
\( = -\frac{1}{2} [0 - \log 5] \)
\( = \frac{1}{2} \log 5 \) The negative sign cancels out when we substitute the values.
In simple words: When you integrate \( \frac{1}{\text{linear expression}} \), you get a logarithm. Don't forget to divide by the coefficient of x. Then, substitute the top and bottom numbers and subtract.

🎯 Exam Tip: Always remember the absolute value sign \( | \ | \) inside the logarithm for indefinite integrals, but when evaluating definite integrals, the absolute value ensures the argument is positive for real logarithms. Pay attention to the sign of the coefficient of x in the denominator.

Question 3.
(i) Let \( I = \int_0^{\pi / 4} \cos^2 3x d x \)
(ii) \( \int_0^{\pi / 4} \tan^2 x dx \)
(iii) \( \int_{\pi / 3}^{\pi / 4}(\tan x+\cot x)^2 dx \)
(iv) \( \int_0^\pi \frac{d x}{1+\sin x} \)
(v) \( \int_0^{\pi / 4} \sin 3 x \sin 2 x d x \)
(vi) \( \int_0^{\pi / 4} \sqrt{1-\sin 2 x} d x \)
Answer:
(i) Let \( I = \int_0^{\pi / 4} \cos^2 3x d x \).
We use the identity \( \cos^2 \theta = \frac{1+\cos 2\theta}{2} \). Here \( \theta = 3x \), so \( 2\theta = 6x \).
\( I = \int_0^{\pi / 4} \frac{1+\cos 6x}{2} d x \)
\( = \frac{1}{2} \int_0^{\pi / 4} (1+\cos 6x) d x \)
\( = \frac{1}{2} [x + \frac{\sin 6x}{6}]_0^{\pi / 4} \)
Apply the limits:
\( = \frac{1}{2} [(\frac{\pi}{4} + \frac{\sin (6 \times \frac{\pi}{4})}{6}) - (0 + \frac{\sin (0)}{6})] \)
\( = \frac{1}{2} [(\frac{\pi}{4} + \frac{\sin (\frac{3\pi}{2})}{6}) - 0] \)
\( = \frac{1}{2} [\frac{\pi}{4} + \frac{-1}{6}] \)
\( = \frac{1}{2} (\frac{\pi}{4} - \frac{1}{6}) \)
\( = \frac{\pi}{8} - \frac{1}{12} \) Using this formula helps to integrate even powers of cosine.
(ii) We need to evaluate \( \int_0^{\pi / 4} \tan^2 x d x \).
We use the identity \( \tan^2 x = \sec^2 x - 1 \).
\( = \int_0^{\pi / 4} (\sec^2 x - 1) d x \)
\( = [\tan x - x]_0^{\pi / 4} \)
Apply the limits:
\( = (\tan (\frac{\pi}{4}) - \frac{\pi}{4}) - (\tan 0 - 0) \)
\( = (1 - \frac{\pi}{4}) - (0 - 0) \)
\( = 1 - \frac{\pi}{4} \) This identity helps simplify the integral into basic forms.
(iii) We need to evaluate \( \int_{\pi / 3}^{\pi / 4}(\tan x+\cot x)^2 dx \).
Expand the square:
\( = \int_{\pi / 3}^{\pi / 4} (\tan^2 x + \cot^2 x + 2 \tan x \cot x) d x \)
Since \( \tan x \cot x = 1 \):
\( = \int_{\pi / 3}^{\pi / 4} (\tan^2 x + \cot^2 x + 2) d x \)
Now, use the identities \( \tan^2 x = \sec^2 x - 1 \) and \( \cot^2 x = \operatorname{cosec}^2 x - 1 \):
\( = \int_{\pi / 3}^{\pi / 4} (\sec^2 x - 1 + \operatorname{cosec}^2 x - 1 + 2) d x \)
\( = \int_{\pi / 3}^{\pi / 4} (\sec^2 x + \operatorname{cosec}^2 x) d x \)
\( = [\tan x - \cot x]_{\pi / 3}^{\pi / 4} \)
Apply the limits:
\( = (\tan (\frac{\pi}{4}) - \cot (\frac{\pi}{4})) - (\tan (\frac{\pi}{3}) - \cot (\frac{\pi}{3})) \)
\( = (1 - 1) - (\sqrt{3} - \frac{1}{\sqrt{3}}) \)
\( = 0 - (\frac{3-1}{\sqrt{3}}) \)
\( = -\frac{2}{\sqrt{3}} \) This problem shows how expanding and using identities can simplify complex integrals.
(iv) We need to evaluate \( \int_0^\pi \frac{d x}{1+\sin x} \).
Multiply the numerator and denominator by \( (1-\sin x) \):
\( = \int_0^\pi \frac{1-\sin x}{(1+\sin x)(1-\sin x)} d x \)
\( = \int_0^\pi \frac{1-\sin x}{1-\sin^2 x} d x \)
We know \( 1-\sin^2 x = \cos^2 x \):
\( = \int_0^\pi \frac{1-\sin x}{\cos^2 x} d x \)
\( = \int_0^\pi (\frac{1}{\cos^2 x} - \frac{\sin x}{\cos^2 x}) d x \)
\( = \int_0^\pi (\sec^2 x - \sec x \tan x) d x \)
\( = [\tan x - \sec x]_0^\pi \)
Apply the limits:
\( = (\tan \pi - \sec \pi) - (\tan 0 - \sec 0) \)
\( = (0 - (-1)) - (0 - 1) \)
\( = 1 - (-1) \)
\( = 2 \) Rationalizing the denominator is a key step here.
(v) We need to evaluate \( \int_0^{\pi / 4} \sin 3x \sin 2x d x \).
Use the product-to-sum identity: \( 2 \sin A \sin B = \cos(A-B) - \cos(A+B) \).
So, \( \sin 3x \sin 2x = \frac{1}{2} (\cos(3x-2x) - \cos(3x+2x)) = \frac{1}{2} (\cos x - \cos 5x) \).
\( = \int_0^{\pi / 4} \frac{1}{2} (\cos x - \cos 5x) d x \)
\( = \frac{1}{2} [\sin x - \frac{\sin 5x}{5}]_0^{\pi / 4} \)
Apply the limits:
\( = \frac{1}{2} [(\sin (\frac{\pi}{4}) - \frac{\sin (5 \times \frac{\pi}{4})}{5}) - (\sin 0 - \frac{\sin 0}{5})] \)
\( = \frac{1}{2} [(\frac{1}{\sqrt{2}} - \frac{\sin (\pi + \frac{\pi}{4})}{5}) - (0 - 0)] \)
\( = \frac{1}{2} [(\frac{1}{\sqrt{2}} - \frac{-\sin (\frac{\pi}{4})}{5})] \)
\( = \frac{1}{2} [(\frac{1}{\sqrt{2}} + \frac{1}{5\sqrt{2}})] \)
\( = \frac{1}{2} [\frac{5+1}{5\sqrt{2}}] \)
\( = \frac{1}{2} \frac{6}{5\sqrt{2}} \)
\( = \frac{3}{5\sqrt{2}} \) Product-to-sum formulas are very useful for integrating products of trigonometric functions.
(vi) We need to evaluate \( \int_0^{\pi / 4} \sqrt{1-\sin 2 x} d x \).
We know that \( 1 = \sin^2 x + \cos^2 x \) and \( \sin 2x = 2 \sin x \cos x \).
So, \( 1-\sin 2x = \sin^2 x + \cos^2 x - 2 \sin x \cos x = (\cos x - \sin x)^2 \).
\( = \int_0^{\pi / 4} \sqrt{(\cos x - \sin x)^2} d x \)
\( = \int_0^{\pi / 4} |\cos x - \sin x| d x \)
For \( 0 < x < \frac{\pi}{4} \), we know that \( \cos x > \sin x \), so \( \cos x - \sin x > 0 \).
Thus, \( |\cos x - \sin x| = \cos x - \sin x \).
\( = \int_0^{\pi / 4} (\cos x - \sin x) d x \)
\( = [\sin x + \cos x]_0^{\pi / 4} \)
Apply the limits:
\( = (\sin (\frac{\pi}{4}) + \cos (\frac{\pi}{4})) - (\sin 0 + \cos 0) \)
\( = (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) - (0 + 1) \)
\( = (\frac{2}{\sqrt{2}}) - 1 \)
\( = \sqrt{2} - 1 \) Recognizing perfect squares under the square root simplifies the problem greatly.
In simple words: For each integral, first simplify the expression using known math rules or identities. Then, find the integral (the reverse of differentiation). Finally, put in the top number (limit) and subtract the result of putting in the bottom number. Always be careful with the signs and fractions.

🎯 Exam Tip: When dealing with definite integrals involving trigonometric functions, look for opportunities to use power-reducing formulas (e.g., for \( \cos^2 x \)), product-to-sum formulas, or fundamental identities to simplify the integrand into a form that is easier to integrate. Pay close attention to the limits of integration for \( |\cos x - \sin x| \) to determine the correct sign.

Question 5.
Evaluate :
(i) \( \int_0^{\pi / 4} \cos^2 3x d x \)
(ii) \( \int_0^{\pi / 4} \tan^2 x dx \)
(iii) \( \int_{\pi / 3}^{\pi / 4}(\tan x+\cot x)^2 d x \)
(iv) \( \int_0^\pi \frac{d x}{1+\sin x} \)
Answer:
(i) Let \( I = \int_0^{\pi / 4} \cos^2 3x d x \).
We use the identity \( \cos^2 \theta = \frac{1+\cos 2\theta}{2} \). Here \( \theta = 3x \), so \( 2\theta = 6x \).
\( I = \int_0^{\pi / 4} \frac{1+\cos 6x}{2} d x \)
\( = \frac{1}{2} \int_0^{\pi / 4} (1+\cos 6x) d x \)
\( = \frac{1}{2} [x + \frac{\sin 6x}{6}]_0^{\pi / 4} \)
Apply the limits:
\( = \frac{1}{2} [(\frac{\pi}{4} + \frac{\sin (6 \times \frac{\pi}{4})}{6}) - (0 + \frac{\sin (0)}{6})] \)
\( = \frac{1}{2} [(\frac{\pi}{4} + \frac{\sin (\frac{3\pi}{2})}{6}) - 0] \)
\( = \frac{1}{2} [(\frac{\pi}{4} + \frac{-1}{6})] \)
\( = \frac{1}{2} (\frac{\pi}{4} - \frac{1}{6}) \)
\( = \frac{\pi}{8} - \frac{1}{12} \) This method helps convert a squared trigonometric term into an integrable form.
(ii) We need to evaluate \( \int_0^{\pi / 4} \tan^2 x d x \).
We use the identity \( \tan^2 x = \sec^2 x - 1 \).
\( = \int_0^{\pi / 4} (\sec^2 x - 1) d x \)
\( = [\tan x - x]_0^{\pi / 4} \)
Apply the limits:
\( = (\tan (\frac{\pi}{4}) - \frac{\pi}{4}) - (\tan 0 - 0) \)
\( = (1 - \frac{\pi}{4}) - (0 - 0) \)
\( = 1 - \frac{\pi}{4} \) The direct integral of \( \tan^2 x \) is not standard, so conversion is necessary.
(iii) We need to evaluate \( \int_{\pi / 3}^{\pi / 4}(\tan x+\cot x)^2 dx \).
Expand the square:
\( = \int_{\pi / 3}^{\pi / 4} (\tan^2 x + \cot^2 x + 2 \tan x \cot x) d x \)
Since \( \tan x \cot x = 1 \):
\( = \int_{\pi / 3}^{\pi / 4} (\tan^2 x + \cot^2 x + 2) d x \)
Now, use the identities \( \tan^2 x = \sec^2 x - 1 \) and \( \cot^2 x = \operatorname{cosec}^2 x - 1 \):
\( = \int_{\pi / 3}^{\pi / 4} (\sec^2 x - 1 + \operatorname{cosec}^2 x - 1 + 2) d x \)
\( = \int_{\pi / 3}^{\pi / 4} (\sec^2 x + \operatorname{cosec}^2 x) d x \)
\( = [\tan x - \cot x]_{\pi / 3}^{\pi / 4} \)
Apply the limits:
\( = (\tan (\frac{\pi}{4}) - \cot (\frac{\pi}{4})) - (\tan (\frac{\pi}{3}) - \cot (\frac{\pi}{3})) \)
\( = (1 - 1) - (\sqrt{3} - \frac{1}{\sqrt{3}}) \)
\( = 0 - (\frac{3-1}{\sqrt{3}}) \)
\( = -\frac{2}{\sqrt{3}} \) Remember to combine the constant terms after using the identities.
(iv) We need to evaluate \( \int_0^\pi \frac{d x}{1+\sin x} \).
Multiply the numerator and denominator by \( (1-\sin x) \):
\( = \int_0^\pi \frac{1-\sin x}{(1+\sin x)(1-\sin x)} d x \)
\( = \int_0^\pi \frac{1-\sin x}{1-\sin^2 x} d x \)
We know \( 1-\sin^2 x = \cos^2 x \):
\( = \int_0^\pi \frac{1-\sin x}{\cos^2 x} d x \)
\( = \int_0^\pi (\frac{1}{\cos^2 x} - \frac{\sin x}{\cos^2 x}) d x \)
\( = \int_0^\pi (\sec^2 x - \sec x \tan x) d x \)
\( = [\tan x - \sec x]_0^\pi \)
Apply the limits:
\( = (\tan \pi - \sec \pi) - (\tan 0 - \sec 0) \)
\( = (0 - (-1)) - (0 - 1) \)
\( = 1 - (-1) \)
\( = 2 \) This technique of multiplying by the conjugate is common for denominators involving sine or cosine.
In simple words: First, simplify the integral using known math rules or identities, especially for squared trigonometric functions. Then, find the anti-derivative for each term. Finally, put in the top limit number and subtract the result from putting in the bottom limit number to get your final answer.

🎯 Exam Tip: Always look for ways to simplify the integrand using trigonometric identities like \( \cos^2 \theta = \frac{1+\cos 2\theta}{2} \) or \( \tan^2 x = \sec^2 x - 1 \). For rational functions with \( (1 \pm \sin x) \) or \( (1 \pm \cos x) \) in the denominator, multiplying by the conjugate is often effective.

Question 6.
Prove that:
(i) \( \int_1^3 \frac{1}{x^2(x+1)} d x=\frac{2}{3}+\log \frac{2}{3} \)
(ii) \( \int_0^a \frac{d x}{\sqrt{5 x-6-x^2}} = \pi \)
(iii) \( \int_1^3 \frac{\log x d x}{(1+x)^2}=\frac{3}{4} \log 3-\log 2 \)
Answer:
(i) Let \( I = \int_1^3 \frac{1}{x^2(x+1)} d x \).
First, we use partial fractions to decompose the integrand:
Let \( \frac{1}{x^2(x+1)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x+1} \).
Multiply both sides by \( x^2(x+1) \):
\( 1 = Ax(x+1) + B(x+1) + Cx^2 \).
When \( x=0 \): \( 1 = B(1) \implies B=1 \).
When \( x=-1 \): \( 1 = C(-1)^2 \implies C=1 \).
When \( x=1 \): \( 1 = A(1)(2) + B(2) + C(1) \implies 1 = 2A + 2B + C \).
Substitute \( B=1 \) and \( C=1 \): \( 1 = 2A + 2(1) + 1 \implies 1 = 2A + 3 \implies 2A = -2 \implies A=-1 \).
So, \( \frac{1}{x^2(x+1)} = -\frac{1}{x} + \frac{1}{x^2} + \frac{1}{x+1} \).
Now integrate:
\( I = \int_1^3 (-\frac{1}{x} + \frac{1}{x^2} + \frac{1}{x+1}) d x \)
\( = [-\log |x| - \frac{1}{x} + \log |x+1|]_1^3 \)
Apply the limits:
\( = (-\log 3 - \frac{1}{3} + \log 4) - (-\log 1 - \frac{1}{1} + \log 2) \)
\( = (-\log 3 - \frac{1}{3} + \log 4) - (0 - 1 + \log 2) \)
\( = -\log 3 - \frac{1}{3} + \log 4 + 1 - \log 2 \)
\( = 1 - \frac{1}{3} + \log 4 - \log 3 - \log 2 \)
\( = \frac{2}{3} + \log (\frac{4}{3 \times 2}) \)
\( = \frac{2}{3} + \log (\frac{2}{3}) \)
This proves the statement. Partial fractions help break down complex rational functions into simpler ones.
(ii) We need to evaluate \( \int_0^a \frac{d x}{\sqrt{5 x-6-x^2}} \). The upper limit 'a' is not explicitly given in the problem statement, but usually it implies the integral covers the full domain where the expression under the square root is positive. Let's complete the square for the quadratic expression under the root:
\( 5x - 6 - x^2 = -(x^2 - 5x + 6) \)
\( = - (x^2 - 5x + (\frac{5}{2})^2 - (\frac{5}{2})^2 + 6) \)
\( = - ((x - \frac{5}{2})^2 - \frac{25}{4} + 6) \)
\( = - ((x - \frac{5}{2})^2 - \frac{25}{4} + \frac{24}{4}) \)
\( = - ((x - \frac{5}{2})^2 - \frac{1}{4}) \)
\( = \frac{1}{4} - (x - \frac{5}{2})^2 \)
So the integral becomes \( \int_0^a \frac{d x}{\sqrt{(\frac{1}{2})^2 - (x - \frac{5}{2})^2}} \).
This is of the form \( \int \frac{dx}{\sqrt{a^2-x^2}} = \sin^{-1} (\frac{x}{a}) + C \).
Here, \( a = \frac{1}{2} \) and \( x \) is \( (x - \frac{5}{2}) \).
The domain for the expression to be real is \( (\frac{1}{2})^2 - (x - \frac{5}{2})^2 \ge 0 \), which means \( (x - \frac{5}{2})^2 \le (\frac{1}{2})^2 \).
\( -\frac{1}{2} \le x - \frac{5}{2} \le \frac{1}{2} \)
\( \frac{5}{2} - \frac{1}{2} \le x \le \frac{5}{2} + \frac{1}{2} \)
\( \frac{4}{2} \le x \le \frac{6}{2} \)
\( 2 \le x \le 3 \).
So, the limits for the integral should be from 2 to 3 for the expression to be defined.
Let's assume 'a' refers to the upper limit of this domain, so we integrate from 2 to 3.
\( = [\sin^{-1} (\frac{x - \frac{5}{2}}{\frac{1}{2}})]_2^3 \)
\( = [\sin^{-1} (2x - 5)]_2^3 \)
Apply the limits:
\( = \sin^{-1} (2(3) - 5) - \sin^{-1} (2(2) - 5) \)
\( = \sin^{-1} (6 - 5) - \sin^{-1} (4 - 5) \)
\( = \sin^{-1} (1) - \sin^{-1} (-1) \)
We know \( \sin^{-1} (1) = \frac{\pi}{2} \) and \( \sin^{-1} (-1) = -\frac{\pi}{2} \).
\( = \frac{\pi}{2} - (-\frac{\pi}{2}) \)
\( = \frac{\pi}{2} + \frac{\pi}{2} \)
\( = \pi \) Completing the square is crucial for integrals involving quadratic expressions under the square root.
(iii) We need to prove \( \int_1^3 \frac{\log x d x}{(1+x)^2}=\frac{3}{4} \log 3-\log 2 \).
We use integration by parts, \( \int u dv = uv - \int v du \).
Let \( u = \log x \) and \( dv = \frac{1}{(1+x)^2} dx \).
Then \( du = \frac{1}{x} dx \) and \( v = \int (1+x)^{-2} dx = -\frac{1}{1+x} \).
So, \( \int_1^3 \frac{\log x}{(1+x)^2} dx = [\log x (-\frac{1}{1+x})]_1^3 - \int_1^3 (-\frac{1}{1+x}) \frac{1}{x} dx \)
\( = [-\frac{\log x}{1+x}]_1^3 + \int_1^3 \frac{1}{x(1+x)} dx \).
First part:
\( = (-\frac{\log 3}{1+3}) - (-\frac{\log 1}{1+1}) \)
\( = -\frac{\log 3}{4} - 0 \) (since \( \log 1 = 0 \)).
Now, for the integral \( \int_1^3 \frac{1}{x(1+x)} dx \). Use partial fractions:
\( \frac{1}{x(1+x)} = \frac{A}{x} + \frac{B}{1+x} \).
\( 1 = A(1+x) + Bx \).
When \( x=0 \): \( 1 = A(1) \implies A=1 \).
When \( x=-1 \): \( 1 = B(-1) \implies B=-1 \).
So, \( \int_1^3 (\frac{1}{x} - \frac{1}{1+x}) d x \).
\( = [\log |x| - \log |1+x|]_1^3 \)
\( = [\log |\frac{x}{1+x}|]_1^3 \)
Apply the limits:
\( = \log (\frac{3}{1+3}) - \log (\frac{1}{1+1}) \)
\( = \log (\frac{3}{4}) - \log (\frac{1}{2}) \)
\( = \log 3 - \log 4 - (\log 1 - \log 2) \)
\( = \log 3 - \log 4 + \log 2 \).
Combine with the first part:
\( = -\frac{\log 3}{4} + \log 3 - \log 4 + \log 2 \)
\( = \frac{3}{4} \log 3 - \log (2^2) + \log 2 \)
\( = \frac{3}{4} \log 3 - 2 \log 2 + \log 2 \)
\( = \frac{3}{4} \log 3 - \log 2 \)
This proves the statement. Integration by parts is a key technique for products of functions.
In simple words: For definite integrals, first solve the integral part. If needed, use tricks like partial fractions or completing the square to simplify it. After integrating, put the top number into the answer, then subtract what you get when you put the bottom number in. Make sure your answer matches the one you need to prove.

🎯 Exam Tip: For proving integral results, methods like partial fractions and integration by parts are frequently used. Remember to correctly choose 'u' and 'dv' for integration by parts (often using the ILATE rule) and to accurately evaluate the limits for both parts of the integral.

Question 7.
Evaluate :
(i) \( \int_0^1 xe^xdx \)
(ii) \( \int_e^{e^2} \frac{d x}{x \log x} \)
(iii) \( \int_1^2\left(\frac{1}{x}-\frac{1}{2 x^2}\right) e^{2 x} d x \)
Answer:
(i) We need to evaluate \( \int_0^1 xe^xdx \).
Use integration by parts: \( \int u dv = uv - \int v du \).
Let \( u = x \) and \( dv = e^x dx \).
Then \( du = dx \) and \( v = e^x \).
\( \int_0^1 xe^x dx = [xe^x]_0^1 - \int_0^1 e^x dx \)
\( = [xe^x - e^x]_0^1 \)
\( = [(x-1)e^x]_0^1 \)
Apply the limits:
\( = ((1-1)e^1) - ((0-1)e^0) \)
\( = (0 \times e) - (-1 \times 1) \)
\( = 0 - (-1) \)
\( = 1 \) This integral is a classic example of integration by parts.
(ii) We need to evaluate \( \int_e^{e^2} \frac{d x}{x \log x} \).
Let \( t = \log x \). Then \( dt = \frac{1}{x} dx \).
Change the limits of integration:
When \( x=e \), \( t = \log e = 1 \).
When \( x=e^2 \), \( t = \log (e^2) = 2 \log e = 2 \).
The integral becomes:
\( = \int_1^2 \frac{dt}{t} \)
\( = [\log |t|]_1^2 \)
Apply the limits:
\( = \log 2 - \log 1 \)
\( = \log 2 - 0 \)
\( = \log 2 \) Substitution is very effective when the derivative of a part of the integrand is also present.
(iii) We need to evaluate \( \int_1^2\left(\frac{1}{x}-\frac{1}{2 x^2}\right) e^{2 x} d x \).
This integral is of the form \( \int e^{ax} (f(x) + f'(x)/a) dx \). Let's try to fit it into the form \( \int e^x (f(x) + f'(x)) dx = e^x f(x) \). This is a trickier form.
Consider the general formula \( \int e^{kx} [f(x) + \frac{1}{k}f'(x)] dx = e^{kx} f(x) \).
Here, \( k=2 \). We want to match \( \frac{1}{x} - \frac{1}{2x^2} \) with \( f(x) + \frac{1}{2}f'(x) \).
Let \( f(x) = \frac{1}{x} \). Then \( f'(x) = -\frac{1}{x^2} \).
So, \( f(x) + \frac{1}{2}f'(x) = \frac{1}{x} + \frac{1}{2}(-\frac{1}{x^2}) = \frac{1}{x} - \frac{1}{2x^2} \).
This matches the integrand perfectly. So the integral is \( [e^{2x} \cdot \frac{1}{x}]_1^2 \).
Apply the limits:
\( = (\frac{e^{2(2)}}{2}) - (\frac{e^{2(1)}}{1}) \)
\( = \frac{e^4}{2} - e^2 \) Recognizing this special form greatly simplifies the integration process.
In simple words: For these integrals, you might use different methods. One method is integration by parts, where you break the problem into two easier parts. Another is substitution, where you replace a tricky part with a new variable. Sometimes, you can spot a special pattern that lets you solve the integral quickly. Always apply the limits by putting the top number in and subtracting the result from the bottom number.

🎯 Exam Tip: Familiarize yourself with special integral forms, especially \( \int e^{kx} [f(x) + \frac{1}{k}f'(x)] dx = e^{kx} f(x) \), as they can save a lot of time. For other integrals, try integration by parts or substitution systematically.

Question 8.
(i) \( \int_0^{\pi / 2} x^2 \sin x dx \)
(ii) \( \int_0^\pi \theta \sin ^2 \theta \cos \theta d \theta \)
Answer:
(i) Let \( I = \int_0^{\pi / 2} x^2 \sin x dx \).
Use integration by parts twice. The ILATE rule suggests \( u=x^2 \) and \( dv=\sin x dx \).
First application:
\( u=x^2 \implies du=2x dx \)
\( dv=\sin x dx \implies v=-\cos x \)
\( I = [x^2 (-\cos x)]_0^{\pi / 2} - \int_0^{\pi / 2} (-\cos x)(2x) dx \)
\( = [-x^2 \cos x]_0^{\pi / 2} + 2 \int_0^{\pi / 2} x \cos x d x \)
Evaluate the first part:
\( = (-(\frac{\pi}{2})^2 \cos (\frac{\pi}{2})) - (-(0)^2 \cos 0) \)
\( = (-(\frac{\pi}{4}) \times 0) - (0 \times 1) = 0 \).
Now, for the second integral, \( \int_0^{\pi / 2} x \cos x d x \). Use integration by parts again.
Let \( u=x \) and \( dv=\cos x dx \).
Then \( du=dx \) and \( v=\sin x \).
\( \int_0^{\pi / 2} x \cos x d x = [x \sin x]_0^{\pi / 2} - \int_0^{\pi / 2} \sin x d x \)
\( = [x \sin x + \cos x]_0^{\pi / 2} \)
Apply limits to this part:
\( = ((\frac{\pi}{2}) \sin (\frac{\pi}{2}) + \cos (\frac{\pi}{2})) - ((0) \sin 0 + \cos 0) \)
\( = ((\frac{\pi}{2}) \times 1 + 0) - (0 + 1) \)
\( = \frac{\pi}{2} - 1 \).
Combine with the first part:
\( I = 0 + 2(\frac{\pi}{2} - 1) = \pi - 2 \) Integrating by parts multiple times is often needed for polynomials multiplied by trigonometric or exponential functions.
(ii) Let \( I = \int_0^\pi \theta \sin^2 \theta \cos \theta d \theta \).
Let \( f(\theta) = \sin \theta \). Then \( f'(\theta) = \cos \theta \).
This integral is of the form \( \int \theta (f(\theta))^2 f'(\theta) d \theta \).
Let \( u = \sin \theta \). Then \( du = \cos \theta d \theta \).
When \( \theta = 0 \), \( u = \sin 0 = 0 \).
When \( \theta = \pi \), \( u = \sin \pi = 0 \).
Since both limits become 0 after substitution, the value of the integral is 0.
\( I = \int_0^0 \theta u^2 du \) Wait, the \( \theta \) term is still there, so simple substitution to 0 limits doesn't immediately yield 0 if \( \theta \) cannot be expressed in terms of \( u \).
Let's re-evaluate the substitution limits. When \( \theta=0, \sin \theta = 0 \). When \( \theta=\pi, \sin \theta = 0 \). So, if we substitute \( u = \sin \theta \), the limits for \( u \) would be from 0 to 0. This means the integral evaluates to 0.
Let's try integration by parts.
Let \( u = \theta \) and \( dv = \sin^2 \theta \cos \theta d \theta \).
Then \( du = d \theta \).
To find \( v \), we integrate \( \int \sin^2 \theta \cos \theta d \theta \). Let \( y = \sin \theta \), so \( dy = \cos \theta d \theta \).
\( \int y^2 dy = \frac{y^3}{3} = \frac{\sin^3 \theta}{3} \). So, \( v = \frac{\sin^3 \theta}{3} \).
\( I = [\theta \frac{\sin^3 \theta}{3}]_0^\pi - \int_0^\pi \frac{\sin^3 \theta}{3} d \theta \).
Evaluate the first part:
\( = (\pi \frac{\sin^3 \pi}{3}) - (0 \frac{\sin^3 0}{3}) \)
\( = (\pi \times 0) - (0 \times 0) = 0 \).
So, \( I = - \frac{1}{3} \int_0^\pi \sin^3 \theta d \theta \).
Use the identity \( \sin 3\theta = 3 \sin \theta - 4 \sin^3 \theta \), so \( \sin^3 \theta = \frac{3 \sin \theta - \sin 3\theta}{4} \).
\( I = -\frac{1}{3} \int_0^\pi \frac{3 \sin \theta - \sin 3\theta}{4} d \theta \)
\( = -\frac{1}{12} \int_0^\pi (3 \sin \theta - \sin 3\theta) d \theta \)
\( = -\frac{1}{12} [-3 \cos \theta + \frac{\cos 3\theta}{3}]_0^\pi \)
Apply the limits:
\( = -\frac{1}{12} [(-3 \cos \pi + \frac{\cos 3\pi}{3}) - (-3 \cos 0 + \frac{\cos 0}{3})] \)
\( = -\frac{1}{12} [(-3(-1) + \frac{-1}{3}) - (-3(1) + \frac{1}{3})] \)
\( = -\frac{1}{12} [(3 - \frac{1}{3}) - (-3 + \frac{1}{3})] \)
\( = -\frac{1}{12} [(\frac{9-1}{3}) - (\frac{-9+1}{3})] \)
\( = -\frac{1}{12} [\frac{8}{3} - \frac{-8}{3}] \)
\( = -\frac{1}{12} [\frac{8}{3} + \frac{8}{3}] \)
\( = -\frac{1}{12} [\frac{16}{3}] \)
\( = -\frac{16}{36} = -\frac{4}{9} \). This type of integral requires careful application of both substitution and integration by parts, along with trigonometric identities.
In simple words: For problems with \( x^2 \) and \( \sin x \), you often have to use a method called 'integration by parts' two times. For other problems, look for ways to substitute parts of the expression or use math rules to change the form of the integral to make it simpler to solve. Always remember to put in the upper and lower limits carefully.

🎯 Exam Tip: When faced with integrals like \( \int x^n \sin x dx \) or \( \int x^n \cos x dx \), expect to use integration by parts \( n \) times. For integrals involving products of trigonometric functions, use product-to-sum identities or substitution. Also, pay attention to potential shortcuts like when the limits of integration become identical after substitution (as initially considered in (ii)).

Question 6. Prove that:
(i) \( \int_1^3 \frac{1}{x^2(x+1)} d x=\frac{2}{3}+\log \frac{2}{3} \)
(ii) \( \int_0^a \frac{d x}{\sqrt{5 x-6-x^2}} = \pi \)
(iii) \( \int_1^3 \frac{\log x d x}{(1+x)^2}=\frac{3}{4} \log 3-\log 2 \)
Answer:
(i) To prove: \( \int_1^3 \frac{1}{x^2(x+1)} d x=\frac{2}{3}+\log \frac{2}{3} \)
Let \( I = \int_1^3 \frac{1}{x^2(x+1)} d x \)
First, we use partial fractions to break down the integrand:
Let \( \frac{1}{x^2(x+1)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x+1} \)
Multiplying both sides by \( x^2(x+1) \):
\( 1 = Ax(x+1) + B(x+1) + Cx^2 \)
Set \( x=0 \): \( 1 = B(0+1) \implies B=1 \)
Set \( x=-1 \): \( 1 = C(-1)^2 \implies C=1 \)
Comparing coefficients of \( x^2 \): \( 0 = A+C \). Since \( C=1 \), \( A=-1 \).
So, \( \frac{1}{x^2(x+1)} = -\frac{1}{x} + \frac{1}{x^2} + \frac{1}{x+1} \)
Now, integrate each term:
\( I = \int_1^3 -\frac{1}{x} d x+\int_1^3 \frac{1}{x^2} d x+\int_1^3 \frac{1}{x+1} d x \)
\( = [-\log|x|]_1^3 + [-\frac{1}{x}]_1^3 + [\log|x+1|]_1^3 \)
\( = (-\log 3 - (-\log 1)) + (-\frac{1}{3} - (-\frac{1}{1})) + (\log|3+1| - \log|1+1|) \)
\( = -\log 3 - 0 + (-\frac{1}{3} + 1) + (\log 4 - \log 2) \)
\( = -\log 3 + \frac{2}{3} + \log(\frac{4}{2}) \)
\( = -\log 3 + \frac{2}{3} + \log 2 \)
\( = \frac{2}{3} + \log 2 - \log 3 \)
\( = \frac{2}{3} + \log (\frac{2}{3}) \) (Using the logarithm property \( \log a - \log b = \log \frac{a}{b} \))
Thus, L.H.S. = R.H.S., and the statement is proven.
In simple words: To solve this integral, we first split the fraction into simpler parts using partial fractions. Then, we integrate each simple part separately and put in the limits from 1 to 3. After combining the log terms, we get the answer shown in the question.

🎯 Exam Tip: Remember to correctly decompose the rational function into partial fractions; a common mistake is to miss the \( \frac{B}{x^2} \) term if the denominator has a repeated factor.

(ii) To prove: \( \int_0^a \frac{d x}{\sqrt{5 x-6-x^2}} = \pi \)
Let \( I = \int_0^a \frac{d x}{\sqrt{5 x-6-x^2}} \)
First, complete the square in the denominator:
\( 5x-6-x^2 = -(x^2-5x+6) \)
\( = -(x^2-5x + (\frac{5}{2})^2 - (\frac{5}{2})^2 + 6) \)
\( = -((x-\frac{5}{2})^2 - \frac{25}{4} + \frac{24}{4}) \)
\( = -((x-\frac{5}{2})^2 - \frac{1}{4}) \)
\( = \frac{1}{4} - (x-\frac{5}{2})^2 \)
Now substitute this back into the integral:
\( I = \int_2^3 \frac{d x}{\sqrt{\frac{1}{4} - (x-\frac{5}{2})^2}} \) (Note: The problem statement has limits \( 0 \) to \( a \). The solution provided has limits \( 2 \) to \( 3 \), which yields \( \pi \). This implies \( a=3 \) and the lower limit is implicitly \( 2 \) to make the expression positive and defined.)
This integral is of the form \( \int \frac{dy}{\sqrt{k^2-y^2}} = \sin^{-1}(\frac{y}{k}) + C \). Here \( k = \frac{1}{2} \) and \( y = x-\frac{5}{2} \).
\( I = [\sin^{-1} \left(\frac{x-\frac{5}{2}}{\frac{1}{2}}\right)]_2^3 \)
\( = [\sin^{-1} (2x-5)]_2^3 \)
Now apply the limits:
\( = \sin^{-1} (2(3)-5) - \sin^{-1} (2(2)-5) \)
\( = \sin^{-1} (6-5) - \sin^{-1} (4-5) \)
\( = \sin^{-1} (1) - \sin^{-1} (-1) \)
We know \( \sin^{-1}(-x) = -\sin^{-1}x \):
\( = \sin^{-1} (1) + \sin^{-1} (1) \)
\( = \frac{\pi}{2} + \frac{\pi}{2} \)
\( = \pi \)
Thus, L.H.S. = R.H.S., and the statement is proven.
In simple words: First, we change the bottom part of the fraction by completing the square to make it a simpler form. Then we recognize this as a known integral that results in an inverse sine function. After putting in the start and end values, we calculate the final answer as \( \pi \).

🎯 Exam Tip: When dealing with integrals involving square roots of quadratic expressions, completing the square in the denominator is often the key first step to convert it into a standard integral form.

(iii) To prove: \( \int_1^3 \frac{\log x d x}{(1+x)^2}=\frac{3}{4} \log 3-\log 2 \)
Let \( I = \int_1^3 \frac{\log x}{(1+x)^2} d x \)
We use integration by parts, with \( u = \log x \) and \( dv = \frac{1}{(1+x)^2} d x \).
Then \( du = \frac{1}{x} d x \) and \( v = \int (1+x)^{-2} d x = \frac{(1+x)^{-1}}{-1} = -\frac{1}{1+x} \).
Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \):
\( I = \left[-\frac{\log x}{1+x}\right]_1^3 - \int_1^3 \left(-\frac{1}{1+x}\right) \frac{1}{x} d x \)
\( I = \left[-\frac{\log 3}{1+3} - \left(-\frac{\log 1}{1+1}\right)\right] + \int_1^3 \frac{1}{x(1+x)} d x \)
Since \( \log 1 = 0 \), the first term simplifies:
\( I = -\frac{\log 3}{4} + \int_1^3 \frac{1}{x(1+x)} d x \)
Now, for the remaining integral, use partial fractions again:
\( \frac{1}{x(1+x)} = \frac{A}{x} + \frac{B}{1+x} \)
\( 1 = A(1+x) + Bx \)
Set \( x=0 \): \( 1 = A(1) \implies A=1 \)
Set \( x=-1 \): \( 1 = B(-1) \implies B=-1 \)
So, \( \int_1^3 \frac{1}{x(1+x)} d x = \int_1^3 \left(\frac{1}{x} - \frac{1}{1+x}\right) d x \)
\( = [\log|x| - \log|1+x|]_1^3 \)
\( = [\log|\frac{x}{1+x}|]_1^3 \)
\( = \log(\frac{3}{1+3}) - \log(\frac{1}{1+1}) \)
\( = \log(\frac{3}{4}) - \log(\frac{1}{2}) \)
Substitute this back into the expression for \( I \):
\( I = -\frac{\log 3}{4} + \log(\frac{3}{4}) - \log(\frac{1}{2}) \)
\( I = -\frac{\log 3}{4} + (\log 3 - \log 4) - (\log 1 - \log 2) \)
\( I = -\frac{\log 3}{4} + \log 3 - \log 4 - 0 + \log 2 \)
\( I = \frac{3}{4}\log 3 - \log 4 + \log 2 \)
Since \( \log 4 = \log (2^2) = 2\log 2 \):
\( I = \frac{3}{4}\log 3 - 2\log 2 + \log 2 \)
\( I = \frac{3}{4}\log 3 - \log 2 \)
Thus, L.H.S. = R.H.S., and the statement is proven.
In simple words: This problem needs two main steps: first, we use integration by parts for the logarithm term. Then, the resulting integral needs to be solved using partial fractions. After evaluating the limits for both parts and simplifying the logarithm expressions, the final answer matches what we needed to prove.

🎯 Exam Tip: Problems requiring integration by parts where one of the resulting terms is a rational function often involve a second step using partial fractions. Always remember that \( \log 1 = 0 \).

Question 7. Evaluate :
(i) \( \int_0^1 xe^xdx \)
(ii) \( \int_e^2 \frac{d x}{x \log x} \)
(iii) \( \int_1^2\left(\frac{1}{x}-\frac{1}{2 x^2}\right) e^{2 x} d x \)
Answer:
(i) Let \( I = \int_0^1 xe^xdx \)
We use integration by parts, with \( u=x \) and \( dv=e^x dx \).
Then \( du=dx \) and \( v=e^x \).
Using the formula \( \int u \, dv = uv - \int v \, du \):
\( I = [xe^x]_0^1 - \int_0^1 e^x dx \)
\( = [xe^x - e^x]_0^1 \)
\( = [(x-1)e^x]_0^1 \)
Now apply the limits:
\( = (1-1)e^1 - (0-1)e^0 \)
\( = (0)e^1 - (-1)e^0 \)
\( = 0 - (-1)(1) \)
\( = 1 \)
In simple words: We use the integration by parts rule to solve this. We choose \( x \) as the first part and \( e^x \) as the second part. After applying the rule and calculating the values at the upper and lower limits, the final answer is 1.

🎯 Exam Tip: For integrals of the form \( \int xe^x dx \), integration by parts is the standard method. Remember that \( e^0 = 1 \).

(ii) Let \( I = \int_e^2 \frac{d x}{x \log x} \)
We use the substitution method. Let \( t = \log x \).
Then, differentiate \( t \) with respect to \( x \): \( \frac{dt}{dx} = \frac{1}{x} \implies dt = \frac{1}{x} dx \).
We also need to change the limits of integration according to the substitution:
When \( x=e \), \( t = \log e = 1 \).
When \( x=e^2 \), \( t = \log(e^2) = 2\log e = 2(1) = 2 \).
Substitute these into the integral:
\( I = \int_1^2 \frac{1}{t} dt \)
\( = [\log|t|]_1^2 \)
Now apply the limits:
\( = \log 2 - \log 1 \)
Since \( \log 1 = 0 \):
\( = \log 2 - 0 \)
\( = \log 2 \)
In simple words: To solve this, we make a substitution where \( \log x \) becomes \( t \). This changes the whole integral into a much simpler form. After finding the integral in terms of \( t \) and putting in the new limits, the answer comes out as \( \log 2 \).

🎯 Exam Tip: Whenever you see a function and its derivative multiplied together (like \( \log x \) and \( 1/x \) here), consider using substitution. Always change the limits of integration when you substitute.

(iii) Let \( I = \int_1^2\left(\frac{1}{x}-\frac{1}{2 x^2}\right) e^{2 x} d x \)
This integral is in the special form \( \int e^{ax} (f(x) + \frac{f'(x)}{a}) dx = \frac{e^{ax} f(x)}{a} \).
Here, \( a=2 \). Let \( f(x) = \frac{1}{x} \).
Then, find the derivative of \( f(x) \): \( f'(x) = -\frac{1}{x^2} \).
Now, check if the integral matches the form: \( e^{2x} (\frac{1}{x} + \frac{-\frac{1}{x^2}}{2}) = e^{2x} (\frac{1}{x} - \frac{1}{2x^2}) \). This matches perfectly.
So, directly apply the formula:
\( I = \left[\frac{e^{2x}}{2} \cdot \frac{1}{x}\right]_1^2 \)
\( = \left[\frac{e^{2x}}{2x}\right]_1^2 \)
Now apply the limits:
\( = \frac{e^{2(2)}}{2(2)} - \frac{e^{2(1)}}{2(1)} \)
\( = \frac{e^4}{4} - \frac{e^2}{2} \)
In simple words: This integral uses a special shortcut formula for expressions with \( e^{ax} \) multiplied by a sum of a function and its derivative. By identifying the function and its derivative, we can directly write the answer in a very quick way, then just put in the upper and lower limits.

🎯 Exam Tip: Recognize the pattern \( \int e^{ax}(f(x) + f'(x)/a) dx = e^{ax}f(x)/a \). This pattern greatly simplifies the solution and is a common trap if not identified.

Question 8.
(i) \( \int_0^{\pi / 2} x^2 \sin x dx \)
(ii) \( \int_0^\pi \theta \sin ^2 \theta \cos \theta d \theta \)
Answer:
(i) Let \( I = \int_0^{\pi/2} x^2 \sin x dx \)
We use integration by parts, \( \int u \, dv = uv - \int v \, du \). We need to apply it twice.
First application: Let \( u=x^2 \) and \( dv=\sin x dx \).
Then \( du=2x dx \) and \( v=-\cos x \).
\( I = [x^2(-\cos x)]_0^{\pi/2} - \int_0^{\pi/2} (-\cos x)(2x) dx \)
\( = [x^2(-\cos x)]_0^{\pi/2} + 2\int_0^{\pi/2} x\cos x dx \)
Evaluate the first term:
\( [x^2(-\cos x)]_0^{\pi/2} = ((\frac{\pi}{2})^2(-\cos\frac{\pi}{2})) - (0^2(-\cos 0)) \)
\( = (\frac{\pi^2}{4} \cdot 0) - (0 \cdot (-1)) = 0 - 0 = 0 \)
So, \( I = 2\int_0^{\pi/2} x\cos x dx \).
Second application of integration by parts for \( \int x\cos x dx \): Let \( u=x \) and \( dv=\cos x dx \).
Then \( du=dx \) and \( v=\sin x \).
\( I = 2 \left( [x\sin x]_0^{\pi/2} - \int_0^{\pi/2} \sin x dx \right) \)
\( = 2 \left( [x\sin x]_0^{\pi/2} - [-\cos x]_0^{\pi/2} \right) \)
Evaluate the terms:
\( [x\sin x]_0^{\pi/2} = (\frac{\pi}{2}\sin\frac{\pi}{2}) - (0\sin 0) = \frac{\pi}{2}(1) - 0 = \frac{\pi}{2} \)
\( [-\cos x]_0^{\pi/2} = (-\cos\frac{\pi}{2}) - (-\cos 0) = (0) - (-1) = 1 \)
Substitute these values back:
\( I = 2 \left( \frac{\pi}{2} - 1 \right) \)
\( I = \pi - 2 \)
In simple words: This problem requires us to use integration by parts two times. First, we separate \( x^2 \) and \( \sin x \), then we do it again for the remaining integral of \( x\cos x \). After applying the limits to all parts, we find the definite integral to be \( \pi - 2 \).

🎯 Exam Tip: For integrals like \( \int x^n \sin x dx \) or \( \int x^n \cos x dx \), repeated application of integration by parts is common. The number of times you apply it usually matches the power of \( x \).

(ii) Let \( I = \int_0^\pi \theta \sin ^2 \theta \cos \theta d \theta \)
First, we need to integrate \( \sin^2\theta \cos\theta \). Let \( t = \sin\theta \). Then \( dt = \cos\theta d\theta \).
So, \( \int \sin^2\theta \cos\theta d\theta = \int t^2 dt = \frac{t^3}{3} = \frac{\sin^3\theta}{3} \).
Now, we use integration by parts for \( I \). Let \( u=\theta \) and \( dv=\sin^2\theta \cos\theta d\theta \).
Then \( du=d\theta \) and \( v=\frac{\sin^3\theta}{3} \).
\( I = \left[\theta \frac{\sin^3\theta}{3}\right]_0^\pi - \int_0^\pi \frac{\sin^3\theta}{3} d\theta \)
Evaluate the first term:
\( \left[\theta \frac{\sin^3\theta}{3}\right]_0^\pi = (\pi \frac{\sin^3\pi}{3}) - (0 \frac{\sin^3 0}{3}) \)
Since \( \sin\pi = 0 \) and \( \sin 0 = 0 \), this term is \( 0 - 0 = 0 \).
So, \( I = -\frac{1}{3} \int_0^\pi \sin^3\theta d\theta \).
Now, use the trigonometric identity \( \sin 3\theta = 3\sin\theta - 4\sin^3\theta \) to express \( \sin^3\theta \):
\( 4\sin^3\theta = 3\sin\theta - \sin 3\theta \implies \sin^3\theta = \frac{1}{4}(3\sin\theta - \sin 3\theta) \).
Substitute this into the integral:
\( I = -\frac{1}{3} \int_0^\pi \frac{1}{4}(3\sin\theta - \sin 3\theta) d\theta \)
\( = -\frac{1}{12} \int_0^\pi (3\sin\theta - \sin 3\theta) d\theta \)
\( = -\frac{1}{12} \left[-3\cos\theta - \left(-\frac{\cos 3\theta}{3}\right)\right]_0^\pi \)
\( = -\frac{1}{12} \left[-3\cos\theta + \frac{\cos 3\theta}{3}\right]_0^\pi \)
Apply the limits:
\( = -\frac{1}{12} \left[ (-3\cos\pi + \frac{\cos 3\pi}{3}) - (-3\cos 0 + \frac{\cos 0}{3}) \right] \)
\( = -\frac{1}{12} \left[ (-3(-1) + \frac{-1}{3}) - (-3(1) + \frac{1}{3}) \right] \)
\( = -\frac{1}{12} \left[ (3 - \frac{1}{3}) - (-3 + \frac{1}{3}) \right] \)
\( = -\frac{1}{12} \left[ (\frac{8}{3}) - (-\frac{8}{3}) \right] \)
\( = -\frac{1}{12} \left[ \frac{8}{3} + \frac{8}{3} \right] \)
\( = -\frac{1}{12} \left[ \frac{16}{3} \right] \)
\( = -\frac{16}{36} = -\frac{4}{9} \)
In simple words: First, we use a substitution to integrate the \( \sin^2\theta \cos\theta \) part. Then, we apply integration by parts. Since a sine term with a power remains, we use a triple-angle identity to simplify it into terms that are easy to integrate. Finally, we put in the limits and calculate the numerical answer.

🎯 Exam Tip: Integrals of products of trigonometric functions often require identities. For powers of sine or cosine, consider double or triple angle formulas to reduce the power for easier integration.

Question 9. Prove that:
(i) \( \int_0^1 \sin ^{-1} \sqrt{x} d x=\frac{\pi}{4} \)
(ii) \( \int_0^1 x^2 \sin ^{-1} x d x=\frac{\pi}{6}-\frac{2}{9} \)
Answer:
(i) To prove: \( \int_0^1 \sin ^{-1} \sqrt{x} d x=\frac{\pi}{4} \)
Let \( I = \int_0^1 \sin^{-1} \sqrt{x} \cdot 1 dx \)
We use integration by parts, \( \int u \, dv = uv - \int v \, du \). Let \( u=\sin^{-1} \sqrt{x} \) and \( dv=1 dx \).
Then \( du = \frac{1}{\sqrt{1-(\sqrt{x})^2}} \cdot \frac{1}{2\sqrt{x}} dx = \frac{1}{2\sqrt{x}\sqrt{1-x}} dx \) and \( v=x \).
\( I = [x\sin^{-1} \sqrt{x}]_0^1 - \int_0^1 x \cdot \frac{1}{2\sqrt{x}\sqrt{1-x}} dx \)
Evaluate the first term:
\( [x\sin^{-1} \sqrt{x}]_0^1 = (1\sin^{-1}\sqrt{1}) - (0\sin^{-1}\sqrt{0}) \)
\( = \sin^{-1}(1) - 0 = \frac{\pi}{2} \).
So, \( I = \frac{\pi}{2} - \frac{1}{2} \int_0^1 \frac{\sqrt{x}}{\sqrt{1-x}} dx \).
To solve the remaining integral \( \int_0^1 \frac{\sqrt{x}}{\sqrt{1-x}} dx \), we use a substitution. Let \( x=\sin^2\theta \).
Then \( dx=2\sin\theta\cos\theta d\theta \).
Limits: when \( x=0, \theta=0 \). When \( x=1, \theta=\frac{\pi}{2} \).
\( \int_0^{\pi/2} \frac{\sqrt{\sin^2\theta}}{\sqrt{1-\sin^2\theta}} (2\sin\theta\cos\theta) d\theta = \int_0^{\pi/2} \frac{\sin\theta}{\cos\theta} (2\sin\theta\cos\theta) d\theta \)
\( = \int_0^{\pi/2} 2\sin^2\theta d\theta \)
Using the identity \( \sin^2\theta = \frac{1-\cos 2\theta}{2} \):
\( = \int_0^{\pi/2} 2 \left(\frac{1-\cos 2\theta}{2}\right) d\theta = \int_0^{\pi/2} (1-\cos 2\theta) d\theta \)
\( = \left[\theta - \frac{\sin 2\theta}{2}\right]_0^{\pi/2} \)
\( = \left(\frac{\pi}{2} - \frac{\sin \pi}{2}\right) - \left(0 - \frac{\sin 0}{2}\right) \)
\( = (\frac{\pi}{2} - 0) - (0 - 0) = \frac{\pi}{2} \).
Substitute this back into the expression for \( I \):
\( I = \frac{\pi}{2} - \frac{1}{2} \left(\frac{\pi}{2}\right) \)
\( = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4} \).
Thus, L.H.S. = R.H.S., and the statement is proven.
In simple words: We start by using integration by parts. This leaves a new integral that needs a substitution. By letting \( x \) be \( \sin^2\theta \), the integral simplifies into a basic trigonometric form. Integrating that and applying the limits, along with the first part of the integration by parts, gives us the required answer of \( \frac{\pi}{4} \).

🎯 Exam Tip: When \( \sin^{-1}\sqrt{x} \) appears, consider the substitution \( x=\sin^2\theta \). This substitution often simplifies the integrand and the limits beautifully.

(ii) To prove: \( \int_0^1 x^2 \sin ^{-1} x d x=\frac{\pi}{6}-\frac{2}{9} \)
Let \( I = \int_0^1 x^2 \sin^{-1} x d x \).
We use the substitution method. Let \( \theta = \sin^{-1} x \).
Then \( x = \sin\theta \), and \( dx = \cos\theta d\theta \).
Limits: when \( x=0, \theta=\sin^{-1} 0 = 0 \). When \( x=1, \theta=\sin^{-1} 1 = \frac{\pi}{2} \).
Substitute these into the integral:
\( I = \int_0^{\pi/2} (\sin\theta)^2 \cdot \theta \cdot \cos\theta d\theta \)
\( I = \int_0^{\pi/2} \theta \sin^2\theta \cos\theta d\theta \)
This integral is similar to Question 8(ii). We solve it using integration by parts.
First, integrate \( \sin^2\theta \cos\theta \). Let \( t = \sin\theta \), then \( dt = \cos\theta d\theta \). So \( \int \sin^2\theta \cos\theta d\theta = \int t^2 dt = \frac{t^3}{3} = \frac{\sin^3\theta}{3} \).
Now, apply integration by parts with \( u=\theta \) and \( dv=\sin^2\theta \cos\theta d\theta \).
Then \( du=d\theta \) and \( v=\frac{\sin^3\theta}{3} \).
\( I = \left[\theta \frac{\sin^3\theta}{3}\right]_0^{\pi/2} - \int_0^{\pi/2} \frac{\sin^3\theta}{3} d\theta \)
Evaluate the first term:
\( \left[\theta \frac{\sin^3\theta}{3}\right]_0^{\pi/2} = \left(\frac{\pi}{2} \cdot \frac{\sin^3(\pi/2)}{3}\right) - \left(0 \cdot \frac{\sin^3 0}{3}\right) \)
\( = \frac{\pi}{2} \cdot \frac{1^3}{3} - 0 = \frac{\pi}{6} \).
Now evaluate the integral \( -\frac{1}{3} \int_0^{\pi/2} \sin^3\theta d\theta \). Use the identity \( \sin^3\theta = \frac{1}{4}(3\sin\theta - \sin 3\theta) \).
\( = -\frac{1}{3} \int_0^{\pi/2} \frac{1}{4}(3\sin\theta - \sin 3\theta) d\theta \)
\( = -\frac{1}{12} \left[-3\cos\theta - \left(-\frac{\cos 3\theta}{3}\right)\right]_0^{\pi/2} \)
\( = -\frac{1}{12} \left[-3\cos\theta + \frac{\cos 3\theta}{3}\right]_0^{\pi/2} \)
Apply the limits:
\( = -\frac{1}{12} \left[ \left(-3\cos\frac{\pi}{2} + \frac{\cos(3\pi/2)}{3}\right) - \left(-3\cos 0 + \frac{\cos 0}{3}\right) \right] \)
\( = -\frac{1}{12} \left[ (-3(0) + \frac{0}{3}) - (-3(1) + \frac{1}{3}) \right] \)
\( = -\frac{1}{12} \left[ 0 - (-3 + \frac{1}{3}) \right] \)
\( = -\frac{1}{12} \left[ -(-\frac{8}{3}) \right] = -\frac{1}{12} \left[ \frac{8}{3} \right] = -\frac{8}{36} = -\frac{2}{9} \).
Combine the terms:
\( I = \frac{\pi}{6} - \frac{2}{9} \).
Thus, L.H.S. = R.H.S., and the statement is proven.
In simple words: We first change the variable from \( x \) to \( \theta \) using \( \theta = \sin^{-1} x \). This transforms the integral into a form that can be solved by integration by parts. The integral of \( \sin^3\theta \) is then found using a trigonometric identity. Adding the results from both parts gives the final answer.

🎯 Exam Tip: Substitution often simplifies inverse trigonometric functions in integrals. Recognize when an integral becomes simpler with a `sin(theta)` substitution for `x`.

Question 10.
(i) If \( \int_0^a 3 x^2dx = 8 \), find the value of a.
(ii) If \( \int_a^b x^3 dx = 0 \) and \( \int_a^b x^2 dx = \frac{2}{3} \), find both a and b.
(iii) If f(x) is of the form f(x) = a + bx + cx², show that \( \int_0^1 f(x) d x=\frac{1}{6}\left[f(0)+4 f\left(\frac{1}{2}\right)+f(1)\right] \)
(iv) if \( \int_0^k \frac{d x}{2+8 x^2}=\frac{\pi}{16} \), find the value of k.
Answer:
(i) Given: \( \int_0^a 3 x^2dx = 8 \)
Integrate the expression:
\( \left[3 \frac{x^{2+1}}{2+1}\right]_0^a = 8 \)
\( \left[3 \frac{x^3}{3}\right]_0^a = 8 \)
\( [x^3]_0^a = 8 \)
Apply the limits:
\( a^3 - 0^3 = 8 \)
\( a^3 = 8 \)
Take the cube root of both sides:
\( a = \sqrt[3]{8} \)
\( a = 2 \)
The quadratic factor \( (a^2+2a+4) \) in the expansion of \( a^3-8 \) gives no real solutions for \( a \).
In simple words: We first find the antiderivative of \( 3x^2 \). Then we put in the upper limit \( a \) and the lower limit \( 0 \), and set the result equal to 8. Solving this simple equation gives us the value of \( a \).

🎯 Exam Tip: Remember that \( \int x^n dx = \frac{x^{n+1}}{n+1} \). When solving for a variable in the limit, carry out the integration and then solve the resulting algebraic equation carefully.

(ii) Given: \( \int_a^b x^3 dx = 0 \) and \( \int_a^b x^2 dx = \frac{2}{3} \). Find \( a \) and \( b \).
From the first condition: \( \int_a^b x^3 dx = 0 \)
\( \left[\frac{x^4}{4}\right]_a^b = 0 \)
\( \frac{b^4}{4} - \frac{a^4}{4} = 0 \)
\( b^4 - a^4 = 0 \)
This can be factored as \( (b^2 - a^2)(b^2 + a^2) = 0 \)
Further factoring: \( (b-a)(b+a)(b^2+a^2) = 0 \)
Since \( a \) and \( b \) are real numbers, \( b^2+a^2 \) is only zero if \( a=0 \) and \( b=0 \), in which case the interval of integration is zero length, and the integral is zero. However, this would contradict the second integral. So \( b^2+a^2 \neq 0 \).
Thus, we have two possibilities:
1. \( b-a = 0 \implies b=a \)
2. \( b+a = 0 \implies b=-a \)

Now, let's use the second condition: \( \int_a^b x^2 dx = \frac{2}{3} \)
\( \left[\frac{x^3}{3}\right]_a^b = \frac{2}{3} \)
\( \frac{b^3}{3} - \frac{a^3}{3} = \frac{2}{3} \)
\( b^3 - a^3 = 2 \)

Case 1: If \( b=a \)
Substitute \( b=a \) into \( b^3 - a^3 = 2 \):
\( a^3 - a^3 = 2 \implies 0 = 2 \). This is a contradiction, so \( b \neq a \).

Case 2: If \( b=-a \)
Substitute \( b=-a \) into \( b^3 - a^3 = 2 \):
\( (-a)^3 - a^3 = 2 \)
\( -a^3 - a^3 = 2 \)
\( -2a^3 = 2 \)
\( a^3 = -1 \)
Take the cube root:
\( a = -1 \)
Since \( b=-a \), then \( b = -(-1) = 1 \).
So, \( a=-1 \) and \( b=1 \).
In simple words: We have two equations from two definite integrals. First, we solve the integral \( \int x^3 dx = 0 \) to find possible relationships between \( a \) and \( b \). Then, we use the integral \( \int x^2 dx = \frac{2}{3} \) to test these relationships and find the specific values for \( a \) and \( b \) that satisfy both conditions.

🎯 Exam Tip: When given multiple conditions involving variables as limits, solve each condition to derive possible relationships. Then, combine these relationships to find the unique solution for the variables.

(iii) Given: \( f(x) = a + bx + cx^2 \). Show that \( \int_0^1 f(x) d x=\frac{1}{6}\left[f(0)+4 f\left(\frac{1}{2}\right)+f(1)\right] \)
First, calculate the Left Hand Side (L.H.S.):
\( L.H.S. = \int_0^1 f(x)dx = \int_0^1 (a+bx+cx^2)dx \)
\( = \left[ax + \frac{bx^2}{2} + \frac{cx^3}{3}\right]_0^1 \)
Apply the limits:
\( = \left(a(1) + \frac{b(1)^2}{2} + \frac{c(1)^3}{3}\right) - \left(a(0) + \frac{b(0)^2}{2} + \frac{c(0)^3}{3}\right) \)
\( = a + \frac{b}{2} + \frac{c}{3} \).

Now, calculate the Right Hand Side (R.H.S.):
First, find the values of \( f(0) \), \( f(\frac{1}{2}) \), and \( f(1) \):
\( f(0) = a + b(0) + c(0)^2 = a \)
\( f(\frac{1}{2}) = a + b(\frac{1}{2}) + c(\frac{1}{2})^2 = a + \frac{b}{2} + \frac{c}{4} \)
\( f(1) = a + b(1) + c(1)^2 = a + b + c \)
Substitute these into the R.H.S. expression:
\( R.H.S. = \frac{1}{6}\left[f(0)+4 f\left(\frac{1}{2}\right)+f(1)\right] \)
\( = \frac{1}{6}\left[a + 4\left(a + \frac{b}{2} + \frac{c}{4}\right) + (a + b + c)\right] \)
\( = \frac{1}{6}\left[a + 4a + 4\left(\frac{b}{2}\right) + 4\left(\frac{c}{4}\right) + a + b + c\right] \)
\( = \frac{1}{6}\left[a + 4a + 2b + c + a + b + c\right] \)
Combine like terms:
\( = \frac{1}{6}\left[6a + 3b + 2c\right] \)
Divide each term by 6:
\( = \frac{6a}{6} + \frac{3b}{6} + \frac{2c}{6} \)
\( = a + \frac{b}{2} + \frac{c}{3} \).

Since L.H.S. \( = a + \frac{b}{2} + \frac{c}{3} \) and R.H.S. \( = a + \frac{b}{2} + \frac{c}{3} \), we have L.H.S. = R.H.S.
Thus, the statement is proven.
In simple words: We calculate both sides of the given equation separately. For the left side, we directly integrate the function \( f(x) \). For the right side, we calculate \( f(x) \) at three specific points and then combine them as instructed. Since both calculations result in the same expression, the proof is complete. This relationship is also known as Simpson's Rule for numerical integration.

🎯 Exam Tip: This problem demonstrates Simpson's 1/3 rule for a quadratic function. For such proof-based questions, carefully calculate both sides of the identity and ensure every algebraic step is correct.

(iv) Given: \( \int_0^k \frac{d x}{2+8 x^2}=\frac{\pi}{16} \). Find the value of k.
First, simplify the denominator of the integrand:
\( \int_0^k \frac{d x}{2(1+4 x^2)}=\frac{\pi}{16} \)
Take the constant out of the integral:
\( \frac{1}{2} \int_0^k \frac{d x}{1+4 x^2}=\frac{\pi}{16} \)
The integral is of the form \( \int \frac{dy}{a^2+y^2} = \frac{1}{a}\tan^{-1}(\frac{y}{a}) \). Here, \( a=1 \). Let \( y=2x \). Then \( dy=2dx \).
So, \( \int \frac{dx}{1+(2x)^2} = \frac{1}{2}\tan^{-1}(2x) \).
Substitute this back:
\( \frac{1}{2} \left[\frac{1}{2}\tan^{-1}(2x)\right]_0^k = \frac{\pi}{16} \)
\( \frac{1}{4} [\tan^{-1}(2x)]_0^k = \frac{\pi}{16} \)
Apply the limits:
\( \tan^{-1}(2k) - \tan^{-1}(2(0)) = \frac{\pi}{16} \cdot 4 \)
\( \tan^{-1}(2k) - \tan^{-1}(0) = \frac{\pi}{4} \)
Since \( \tan^{-1}(0)=0 \):
\( \tan^{-1}(2k) = \frac{\pi}{4} \)
To find \( 2k \), take the tangent of both sides:
\( 2k = \tan\left(\frac{\pi}{4}\right) \)
We know that \( \tan(\frac{\pi}{4}) = 1 \):
\( 2k = 1 \)
\( k = \frac{1}{2} \)
In simple words: We first simplify the integral by factoring out a constant from the bottom. Then, we recognize it as a standard integral whose solution involves the inverse tangent function. After applying the limits and doing the calculations, we find the value of \( k \).

🎯 Exam Tip: When dealing with integrals involving \( (a^2+x^2) \) or \( (a^2+b^2x^2) \) in the denominator, remember the inverse tangent integral formula. Always adjust for any coefficient of \( x \) in the argument of the inverse tangent.

Question 10.
(ii) If \( \int_a^b x^2 d x=0 \) and if \( \int_a^b x^2 d x=\frac{2}{3} \) find both a and b.
Answer: We need to find the values of \( a \) and \( b \) using two given conditions.
The problem states the first condition as \( \int_a^b x^2 d x = 0 \). However, the provided solution uses \( \int_a^b x^3 d x = 0 \). Following the solution's steps, we use:
Given \( \int_a^b x^3 d x = 0 \)
\( \implies \) \( \left.\frac{x^4}{4}\right]_a^b = 0 \)
\( \implies \) \( \frac{1}{4}(b^4 - a^4) = 0 \)
\( \implies \) \( (b - a)(b + a)(b^2 - a^2) = 0 \) ... (1) This equation shows that one of its factors must be zero.
The second condition is \( \int_a^b x^2 d x=\frac{2}{3} \)
\( \implies \) \( \left.\frac{x^3}{3}\right]_a^b=\frac{2}{3} \)
\( \implies \) \( \frac{1}{3}\left(b^3-a^3\right)=\frac{2}{3} \)
\( \implies \) \( b^3 - a^3 = 2 \)
\( \implies \) \( (b - a) (b^2 + ab + a^2) = 2 \) ... (2)
From equation (1), if \( b - a = 0 \), then \( b = a \).
Substituting \( b = a \) into equation (2) gives \( (a - a)(a^2 + a \cdot a + a^2) = 2 \), which simplifies to \( 0 = 2 \). This is not possible, so \( b \neq a \).
Next, from equation (1), if \( b + a = 0 \), then \( b = -a \).
Substituting \( b = -a \) into equation (2):
\( (-a - a) ((-a)^2 + a(-a) + a^2) = 2 \)
\( (-2a) (a^2 - a^2 + a^2) = 2 \)
\( (-2a) (a^2) = 2 \)
\( \implies \) \( -2a^3 = 2 \)
\( \implies \) \( a^3 = -1 \)
\( \implies \) \( a = -1 \)
Since \( b = -a \), then \( b = -(-1) = 1 \).
Thus, the values are \( a = -1 \) and \( b = 1 \).
In simple words: We had two mathematical rules involving \( a \) and \( b \). We systematically used these rules to find their values. After checking different possibilities, we found that \( a \) must be -1 and \( b \) must be 1 for both rules to be true.

🎯 Exam Tip: When a question provides multiple conditions, make sure to use all of them in your solution. Always verify that your final values satisfy all given conditions.

Question 10.
(iii) If \( f(x) \) is of the form \( f(x) = a + bx + cx^2 \), show that \( \int_0^1 f(x) d x=\frac{1}{6}\left[f(0)+4 f\left(\frac{1}{2}\right)+f(1)\right] \).
Answer: We need to prove that the left-hand side (L.H.S.) of the equation is equal to the right-hand side (R.H.S.).
Let's calculate the Left-Hand Side (L.H.S.):
L.H.S. \( = \int_0^1 f(x) d x \)
Substitute \( f(x) = a + bx + cx^2 \):
\( = \int_0^1 (a + bx + cx^2) d x \)
Integrate each term:
\( = \left[ax + \frac{bx^2}{2} + \frac{cx^3}{3}\right]_0^1 \)
Apply the limits of integration (from 0 to 1):
\( = \left(a(1) + \frac{b(1)^2}{2} + \frac{c(1)^3}{3}\right) - \left(a(0) + \frac{b(0)^2}{2} + \frac{c(0)^3}{3}\right) \)
\( = a + \frac{b}{2} + \frac{c}{3} \)
So, L.H.S. \( = a + \frac{b}{2} + \frac{c}{3} \).
Now, let's calculate the Right-Hand Side (R.H.S.):
First, find \( f(0) \), \( f\left(\frac{1}{2}\right) \), and \( f(1) \) using \( f(x) = a + bx + cx^2 \).
For \( x = 0 \):
\( f(0) = a + b(0) + c(0)^2 = a \)
For \( x = \frac{1}{2} \):
\( f\left(\frac{1}{2}\right) = a + b\left(\frac{1}{2}\right) + c\left(\frac{1}{2}\right)^2 = a + \frac{b}{2} + \frac{c}{4} \)
For \( x = 1 \):
\( f(1) = a + b(1) + c(1)^2 = a + b + c \)
Substitute these values into the R.H.S. expression:
R.H.S. \( = \frac{1}{6}\left[f(0)+4 f\left(\frac{1}{2}\right)+f(1)\right] \)
\( = \frac{1}{6}\left[a + 4\left(a + \frac{b}{2} + \frac{c}{4}\right) + (a + b + c)\right] \)
Distribute the 4 inside the parenthesis:
\( = \frac{1}{6}\left[a + 4a + \frac{4b}{2} + \frac{4c}{4} + a + b + c\right] \)
\( = \frac{1}{6}\left[a + 4a + 2b + c + a + b + c\right] \)
Combine the like terms:
\( = \frac{1}{6}\left[(1+4+1)a + (2+1)b + (1+1)c\right] \)
\( = \frac{1}{6}\left[6a + 3b + 2c\right] \)
Multiply by \( \frac{1}{6} \):
\( = \frac{6a}{6} + \frac{3b}{6} + \frac{2c}{6} \)
\( = a + \frac{b}{2} + \frac{c}{3} \)
So, R.H.S. \( = a + \frac{b}{2} + \frac{c}{3} \).
Since L.H.S. \( = a + \frac{b}{2} + \frac{c}{3} \) and R.H.S. \( = a + \frac{b}{2} + \frac{c}{3} \),
Thus, L.H.S. = R.H.S. The proof is complete.
In simple words: We showed that both sides of the equation simplify to the same expression. We first calculated the integral on the left side. Then, we calculated the values of \( f(x) \) at 0, \( \frac{1}{2} \), and 1 and put them into the right side. Since both calculations gave the exact same result, the statement is proven true.

🎯 Exam Tip: When proving an identity, it's best to simplify the L.H.S. and R.H.S. independently until they match. Be careful with algebraic calculations and distributing terms.

Question 10.
(iv) if \( \int_0^k \frac{d x}{2+8 x^2}=\frac{\pi}{16} \), find the value of k.
Answer: We need to find the value of \( k \) from the given definite integral equation.
Given: \( \int_0^k \frac{d x}{2+8 x^2}=\frac{\pi}{16} \)
First, simplify the denominator of the integrand by factoring out 8:
\( \int_0^k \frac{d x}{8\left(\frac{2}{8}+x^2\right)}=\frac{\pi}{16} \)
\( \int_0^k \frac{d x}{8\left(\frac{1}{4}+x^2\right)}=\frac{\pi}{16} \)
Now, take the constant \( \frac{1}{8} \) out of the integral:
\( \frac{1}{8} \int_0^k \frac{d x}{x^2+\left(\frac{1}{2}\right)^2}=\frac{\pi}{16} \)
This integral is in the standard form \( \int \frac{d x}{x^2+a^2} = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) \). Here, \( a = \frac{1}{2} \).
Apply the integration formula:
\( \frac{1}{8} \left[\frac{1}{\frac{1}{2}} \tan^{-1}\left(\frac{x}{\frac{1}{2}}\right)\right]_0^k = \frac{\pi}{16} \)
\( \frac{1}{8} \left[2 \tan^{-1}(2x)\right]_0^k = \frac{\pi}{16} \)
Simplify the constant term:
\( \frac{2}{8} \left[\tan^{-1}(2x)\right]_0^k = \frac{\pi}{16} \)
\( \frac{1}{4} \left[\tan^{-1}(2x)\right]_0^k = \frac{\pi}{16} \)
Now, apply the limits of integration from 0 to \( k \):
\( \frac{1}{4} \left[\tan^{-1}(2k) - \tan^{-1}(2 \cdot 0)\right] = \frac{\pi}{16} \)
\( \frac{1}{4} \left[\tan^{-1}(2k) - \tan^{-1}(0)\right] = \frac{\pi}{16} \)
We know that \( \tan^{-1}(0) = 0 \):
\( \frac{1}{4} \left[\tan^{-1}(2k) - 0\right] = \frac{\pi}{16} \)
\( \frac{1}{4} \tan^{-1}(2k) = \frac{\pi}{16} \)
Multiply both sides by 4:
\( \tan^{-1}(2k) = \frac{\pi}{16} \times 4 \)
\( \tan^{-1}(2k) = \frac{\pi}{4} \)
To find \( 2k \), take the tangent of both sides:
\( 2k = \tan\left(\frac{\pi}{4}\right) \)
We know that \( \tan\left(\frac{\pi}{4}\right) = 1 \):
\( 2k = 1 \)
Solve for \( k \):
\( k = \frac{1}{2} \)
In simple words: We were given an equation with an integral and needed to find the value of \( k \). We first simplified the integral expression. Then, we used a known integration formula and put in the given limits. This helped us set up a simpler equation, which we solved to find that \( k \) is \( \frac{1}{2} \).

🎯 Exam Tip: Always look for ways to simplify the integrand before integrating. Recognizing standard integral forms will save time and prevent errors. Remember the values of trigonometric functions for common angles like \( \frac{\pi}{4} \).