CBSE Class 12 Mathematics Integrals Assignment Set 06

Selected NCERT Questions

Question. Write the value of \( \int \sec x(\sec x + \tan x) dx \).
Answer: \( \int \sec x(\sec x + \tan x) dx \)
\( = \int \sec^2 x dx + \int \sec x \tan x dx \)
\( = \tan x + \sec x + C \)
\( \left[ \because \frac{d}{dx} (\tan x) = \sec^2 x \text{ and } \frac{d}{dx} (\sec x) = \sec x \tan x \right] \)

Question. Find: \( \int \frac{e^{2x} - e^{-2x}}{e^{2x} + e^{-2x}} dx \) 
Answer: Put \( (e^{2x} + e^{-2x}) = t \)
\( \implies \) \( (2e^{2x} - 2e^{-2x}) dx = dt \)
\( \implies \) \( (e^{2x} - e^{-2x}) dx = \frac{dt}{2} \)
\( \therefore \int \frac{e^{2x} - e^{-2x}}{e^{2x} + e^{-2x}} dx = \frac{1}{2} \int \frac{1}{t} dt = \frac{1}{2} \log | t | + C = \frac{1}{2} \log | e^{2x} + e^{-2x} | + C \)

Question. Find: \( \int \frac{\sqrt{\tan x}}{\sin x \cos x} dx \)
Answer: \( \int \frac{\sqrt{\tan x}}{\sin x \cos x} dx = \int \frac{\sqrt{\tan x}}{\frac{\sin x}{\cos x} \cdot \cos^2 x} dx = \int \frac{\sec^2 x}{\sqrt{\tan x}} dx \)
Put \( \tan x = t \)
\( \implies \) \( \sec^2 x dx = dt \).
\( \therefore \int \frac{\sec^2 x}{\sqrt{\tan x}} dx = \int \frac{1}{\sqrt{t}} dt = \int t^{-1/2} dt = \frac{t^{1/2}}{1/2} + C = 2\sqrt{t} + C = 2\sqrt{\tan x} + C \)

Question. Find: \( \int \frac{(x + 1)(x + \log x)^2}{x} dx \)
Answer: \( \int \frac{(x + 1)(x + \log x)^2}{x} dx = \int \left( \frac{x + 1}{x} \right)(x + \log x)^2 dx = \int \left( 1 + \frac{1}{x} \right)(x + \log x)^2 dx \)
Put \( x + \log x = t \)
\( \implies \) \( \left( 1 + \frac{1}{x} \right) dx = dt \)
\( \therefore \int \left( 1 + \frac{1}{x} \right)(x + \log x)^2 dx = \int t^2 dt = \frac{t^3}{3} + C = \frac{1}{3}(x + \log x)^3 + C \)

Question. Find: \( \int \frac{x^3 \sin(\tan^{-1} x^4)}{1 + x^8} dx \)
Answer: \( \int \frac{x^3 \sin(\tan^{-1} x^4)}{1 + x^8} dx \), put \( \tan^{-1}(x^4) = t \)
\( \implies \) \( \frac{4x^3}{1 + x^8} dx = dt \)
\( \implies \) \( \frac{x^3}{1 + x^8} dx = \frac{dt}{4} \)
\( \therefore \int \frac{x^3 \sin(\tan^{-1} x^4)}{1 + x^8} dx = \frac{1}{4} \int \sin t \, dt = \frac{1}{4}(-\cos t) + C = \frac{-1}{4} \cos(\tan^{-1} x^4) + C \)

Question. Find: \( \int \frac{\cos 2x - \cos 2\alpha}{\cos x - \cos \alpha} dx \)
Answer: Let \( I = \int \frac{\cos 2x - \cos 2\alpha}{\cos x - \cos \alpha} dx = \int \frac{(2 \cos^2 x - 1) - (2 \cos^2 \alpha - 1)}{\cos x - \cos \alpha} dx = 2 \int \frac{\cos^2 x - \cos^2 \alpha}{\cos x - \cos \alpha} dx \)
\( = 2 \int \frac{(\cos x - \cos \alpha)(\cos x + \cos \alpha)}{(\cos x - \cos \alpha)} dx = 2 \int (\cos x + \cos \alpha) dx \)
\( = 2 \int \cos x \, dx + \cos \alpha \int 1. dx = 2 \sin x + 2x \cos \alpha + C \)

Question. Find: \( \int \frac{1}{\sqrt{x^2 + 2x + 2}} dx \)
Answer: \( \int \frac{1}{\sqrt{x^2 + 2x + 2}} dx = \int \frac{1}{\sqrt{(x^2 + 2x + 1) + 1}} dx = \int \frac{1}{\sqrt{(x + 1)^2 + 1}} dx \)
Putting \( x + 1 = t \)
\( \implies \) \( dx = dt \)
\( \therefore \int \frac{1}{\sqrt{(x + 1)^2 + 1}} dx = \int \frac{1}{\sqrt{(t)^2 + (1)^2}} dt \)
\( = \log | t + \sqrt{t^2 + 1} | + C = \log | (x + 1) + \sqrt{(x^2 + 2x + 1) + 1} | + C \)
\( = \log | (x + 1) + \sqrt{x^2 + 2x + 2} | + C \)

Question. Find: \( \int \frac{1}{9x^2 + 6x + 5} dx \)
Answer: \( \int \frac{1}{9x^2 + 6x + 5} dx = \frac{1}{9} \int \frac{1}{x^2 + \frac{6}{9}x + \frac{5}{9}} dx \)
\( = \frac{1}{9} \int \frac{1}{x^2 + \frac{2}{3}x + \frac{5}{9} + \left(\frac{1}{3}\right)^2 - \left(\frac{1}{3}\right)^2} dx = \frac{1}{9} \int \frac{1}{\left(x + \frac{1}{3}\right)^2 + \left(\frac{2}{3}\right)^2} dx \), putting \( x + \frac{1}{3} = t \)
\( \implies \) \( dx = dt \)
\( = \frac{1}{9} \int \frac{1}{t^2 + \left(\frac{2}{3}\right)^2} dt = \frac{1}{9} \cdot \frac{1}{\frac{2}{3}} \tan^{-1} \left( \frac{t}{2/3} \right) + C = \frac{1}{6} \tan^{-1} \left[ \frac{3\left(x + \frac{1}{3}\right)}{2} \right] + C = \frac{1}{6} \tan^{-1} \left( \frac{3x + 1}{2} \right) + C \)

Question. Evaluate: \( \int \frac{5x + 3}{\sqrt{x^2 + 4x + 10}} dx \)
Answer: We can express the \( N^r \) as \( 5x + 3 = A \frac{d}{dx} (x^2 + 4x + 10) + B \)
\( \implies \) \( 5x + 3 = A(2x + 4) + B \)
\( \implies \) \( 5x + 3 = 2Ax + (4A + B) \)
Equating the coefficients, we get
\( 2A = 5 \quad \text{and} \quad 4A + B = 3 \)
\( A = \frac{5}{2} \)
\( \implies \) \( 4 \times \frac{5}{2} + B = 3 \)
\( \implies \) \( B = 3 - 10 = -7 \)
\( \therefore 5x + 3 = \frac{5}{2}(2x + 4) + (-7) \)
\( \therefore I = \int \frac{\frac{5}{2}(2x + 4) - 7}{\sqrt{x^2 + 4x + 10}} dx = \frac{5}{2} \int \frac{(2x + 4)}{\sqrt{x^2 + 4x + 10}} dx - 7 \int \frac{dx}{\sqrt{x^2 + 4x + 10}} \)
\( I = \frac{5}{2} I_1 - 7 I_2 \quad ... (i) \)
where \( I_1 = \int \frac{2x + 4}{\sqrt{x^2 + 4x + 10}} dx \) and \( I_2 = \int \frac{dx}{\sqrt{x^2 + 4x + 10}} \)
Now, \( I_1 = \int \frac{(2x + 4)}{\sqrt{x^2 + 4x + 10}} dx \)
Let \( x^2 + 4x + 10 = t \)
\( \implies \) \( (2x + 4)dx = dt \)
\( \therefore I_1 = \int \frac{dt}{\sqrt{t}} = \int t^{-1/2} dt = \frac{t^{-1/2 + 1}}{-\frac{1}{2} + 1} + C_1 = 2\sqrt{t} + C_1 \)
\( I_1 = 2\sqrt{x^2 + 4x + 10} + C_1 \)
Again, \( I_2 = \int \frac{dx}{\sqrt{x^2 + 2.x.2 + 2^2 - 4 + 10}} = \int \frac{dx}{\sqrt{(x + 2)^2 + (\sqrt{6})^2}} \)
\( = \log | (x + 2) + \sqrt{x^2 + 4x + 10} | + C_2 \)
Putting the value of \( I_1 \) and \( I_2 \) in (i), we get
\( I = \frac{5}{2} \times 2\sqrt{x^2 + 4x + 10} - 7 \log | (x + 2) + \sqrt{x^2 + 4x + 10} | + \left( \frac{5}{2} C_1 - 7C_2 \right) \)
\( = 5\sqrt{x^2 + 4x + 10} - 7 \log | (x + 2) + \sqrt{x^2 + 4x + 10} | + C \)

Question. Find: \( \int \frac{1 - x^2}{x(1 - 2x)} dx \)
Answer: \( \int \frac{1 - x^2}{x(1 - 2x)} dx = \int \left[ \frac{1}{2} + \frac{-\frac{1}{2}x + 1}{x(1 - 2x)} \right] dx = \frac{1}{2} \int dx - \frac{1}{2} \int \frac{x - 2}{x(1 - 2x)} dx = \frac{x}{2} - \frac{1}{2} I_1 \quad ... (i) \)
Now, \( I_1 = \int \frac{x - 2}{x(1 - 2x)} dx \)
\( \because \frac{x - 2}{x(1 - 2x)} \) is a proper rational function
\( \therefore \frac{x - 2}{x(1 - 2x)} = \frac{A}{x} + \frac{B}{1 - 2x} \quad ... (ii) \)
\( \implies \) \( x - 2 = A(1 - 2x) + Bx \)
\( \implies \) \( x - 2 = (-2A + B)x + A \)
\( \implies \) \( A = -2 \text{ and } -2A + B = 1 \)
\( \implies \) \( B = 1 + 2A = 1 + 2(-2) = -3 \)
Putting values of \( A \) and \( B \) in (ii), we have
\( \frac{x - 2}{x(1 - 2x)} = \frac{-2}{x} - \frac{3}{1 - 2x} \)
\( \therefore \int \frac{x - 2}{x(1 - 2x)} dx = \int \left[ \frac{-2}{x} - \frac{3}{1 - 2x} \right] dx \)
\( = -2 \int \frac{1}{x} dx - 3 \int \frac{1}{1 - 2x} dx = -2 \log|x| - 3 \frac{\log|1 - 2x|}{-2} + C_1 \)
\( = -2 \log|x| + \frac{3}{2} \log|1 - 2x| + C_1 \)
Putting the value \( I_1 \) in (i), we have
\( \int \frac{1 - x^2}{x(1 - 2x)} dx = \frac{x}{2} - \frac{1}{2} \left[ -2 \log|x| + \frac{3}{2} \log|1 - 2x| + C_1 \right] \)
\( = \frac{x}{2} + \log|x| - \frac{3}{4} \log|1 - 2x| - \frac{C_1}{2} \)
\( = \frac{x}{2} + \log|x| - \frac{3}{4} \log|1 - 2x| + C \), where \( C = -\frac{C_1}{2} \)

Question. Find: \( \int \frac{x^3 + x + 1}{x^2 - 1} dx \)
Answer: \( \int \frac{x^3 + x + 1}{x^2 - 1} dx = \int \left[ x + \frac{2x + 1}{x^2 - 1} \right] dx = \int x dx + \int \frac{2x + 1}{x^2 - 1} dx = \frac{x^2}{2} + \int \frac{2x}{x^2 - 1} dx + \int \frac{1}{x^2 - 1} dx \)
Putting \( x^2 - 1 = t \)
\( \implies \) \( 2x dx = dt \) in second integral, we get
\( = \frac{x^2}{2} + \int \frac{1}{t} dt + \int \frac{1}{x^2 - (1)^2} dx = \frac{x^2}{2} + \log|t| + \frac{1}{2(1)} \log \left| \frac{x - 1}{x + 1} \right| + C \)
\( = \frac{x^2}{2} + \log|x^2 - 1| + \frac{1}{2} \log \left| \frac{x - 1}{x + 1} \right| + C \)

Question. Evaluate: \( \int \frac{x \cos^{-1} x}{\sqrt{1 - x^2}} dx \)
Answer: Let \( I = \int \frac{x \cos^{-1} x}{\sqrt{1 - x^2}} dx \)
Put \( \cos^{-1} x = z \)
\( \implies \) \( \frac{-1}{\sqrt{1 - x^2}} dx = dz \)
\( \implies \) \( \frac{1}{\sqrt{1 - x^2}} dx = -dz \)
\( I = - \int \cos z . z dz = - z . \sin z - \int \sin z dz + C = - (z \sin z + \cos z - C) = - z \sin z - \cos z + C \)
\( I = - \cos^{-1} x . \sqrt{1 - x^2} - x + C \qquad \qquad [\because x = \cos z \)
\( \implies \) \( \sin z = \sqrt{1 - x^2}] \)
\( I = -\sqrt{1 - x^2} \cos^{-1} x - x + C \)

Question. Find: \( \int e^x \left( \frac{1 + \sin x}{1 + \cos x} \right) dx \)
Answer: Let \( I = \int e^x \left( \frac{1 + \sin x}{1 + \cos x} \right) dx = \int e^x \left[ \frac{1 + 2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos^2 \frac{x}{2}} \right] dx \)
\( = \int e^x \left[ \frac{1}{2 \cos^2 \frac{x}{2}} + \frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos^2 \frac{x}{2}} \right] dx = \int e^x \left( \frac{1}{2} \sec^2 \frac{x}{2} + \tan \frac{x}{2} \right) dx \)
\( = \int \underset{\text{I}}{e^x} \underset{\text{II}}{\tan \frac{x}{2}} dx + \frac{1}{2} \int e^x \sec^2 \frac{x}{2} dx \)
\( = \left( \tan \frac{x}{2} \right) e^x - \int \left( \frac{1}{2} \sec^2 \frac{x}{2} \right) e^x dx + \frac{1}{2} \int e^x \sec^2 \frac{x}{2} dx \)
\( \implies \) \( I = e^x \tan \frac{x}{2} + C \)

Question. Evaluate: \( \int_1^4 (x^2 - x) dx \) as limit of sums.
Answer: \( \int_1^4 (x^2 - x) dx \)
We have to solve it by using limit of sums.
Here, \( a = 1, b = 4, h = \frac{b - a}{n} = \frac{4 - 1}{n} \) i.e., \( nh = 3 \)
Limit of sum for \( \int_1^4 (x^2 - x) dx \) is
\( = \lim_{h \to 0} h[f(1) + f(1 + h) + f(1 + 2h) + .... + f\{1 + (n - 1)h\}] \)
Now, \( f(1) = 1 - 1 = 0 \)
\( f(1 + h) = (1 + h)^2 - (1 + h) = h^2 + h \)
\( f(1 + 2h) = (1 + 2h)^2 - (1 + 2h) = 4h^2 + 2h \)
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\( f[1 + (n - 1)h] = \{1 + (n - 1)h\}^2 - \{1 + (n - 1)h\} = (n - 1)^2 h^2 + (n - 1)h \)
\( \therefore \int_1^4 (x^2 - x) dx = \lim_{h \to 0} h[0 + h^2 + h + 4h^2 + 2h + .... (n - 1)^2 h^2 + (n - 1)h] \)
\( = \lim_{h \to 0} h[h^2 \{1 + 4 + .. + (n - 1)^2\} + h\{1 + 2 + ... (n - 1)\}] \)
\( = \lim_{h \to 0} h \left[ h^2 . \frac{(n)(n - 1)(2n - 1)}{6} + h \frac{n(n - 1)}{2} \right] \)
\( \left[ \because 1 + 4 + ... + (n - 1)^2 = \frac{n(n - 1)(2n - 1)}{6} \text{ and } 1 + 2 + ... + (n - 1) = \frac{n(n - 1)}{2} \right] \)
\( = \lim_{h \to 0} \left[ \frac{nh(nh - h)(2nh - h)}{6} + \frac{nh(nh - h)}{2} \right] \)
\( = \lim_{h \to 0} \left[ \frac{(3 - h)(3)(6 - h)}{6} + \frac{(3 - h)(3)}{2} \right] = \left( \frac{3 \times 3 \times 6}{6} \right) + \left( \frac{3 \times 3}{2} \right) = 9 + \frac{9}{2} = \frac{27}{2} \)

Question. Evaluate: \( \int_{-5}^5 |x + 2| dx \)
Answer: Here, function is \( |x + 2| \) which is defined as
\( |x + 2| = (x + 2) \), if \( x > -2 \)
\( \qquad \quad - (x + 2) \), if \( x < -2 \)
So, we have
\( \int_{-5}^5 |x + 2| dx = \int_{-5}^{-2} -(x + 2) dx + \int_{-2}^5 (x + 2) dx \qquad \left[ \int_a^b f(x)dx = \int_a^c f(x)dx + \int_c^b f(x)dx \right] \) where \( a < c < b \)
\( = \left[ -\frac{x^2}{2} - 2x \right]_{-5}^{-2} + \left[ \frac{x^2}{2} + 2x \right]_{-2}^5 \)
\( = \frac{-(-2)^2}{2} - 2(-2) + \frac{-(-5)^2}{2} + 2 \times (-5) + \frac{(5)^2}{2} + 2 \times (5) - \frac{(-2)^2}{2} - 2 \times (-2) \)
\( = -2 + 4 + \frac{25}{2} - 10 + \frac{25}{2} + 10 - 2 + 4 = 29 \)

Question. Evaluate: \( \int_0^{\pi/4} \log(1 + \tan x) dx \)
Answer: Let \( I = \int_0^{\pi/4} \log(1 + \tan x) dx \quad ... (i) \)
\( \therefore I = \int_0^{\pi/4} \log \left[ 1 + \tan \left( \frac{\pi}{4} - x \right) \right] dx \quad \text{(By using property } \int_0^a f(x)dx = \int_0^a f(a - x)dx) \)
\( = \int_0^{\pi/4} \log \left[ 1 + \frac{\tan \frac{\pi}{4} - \tan x}{1 + \tan \frac{\pi}{4} . \tan x} \right] dx \)
\( = \int_0^{\pi/4} \log \left[ 1 + \frac{1 - \tan x}{1 + \tan x} \right] dx = \int_0^{\pi/4} \log \left[ \frac{1 + \tan x + 1 - \tan x}{1 + \tan x} \right] dx \)
\( I = \int_0^{\pi/4} \log \frac{2}{1 + \tan x} dx = \int_0^{\pi/4} [\log 2 - \log(1 + \tan x)] dx \quad ... (ii) \)
Adding (i) and (ii), we get
\( 2I = \int_0^{\pi/4} \log 2 \, dx = \log 2 \int_0^{\pi/4} dx = \log 2 [x]_0^{\pi/4} \)
\( 2I = \frac{\pi}{4} \log 2 \)
\( \implies \) \( I = \frac{\pi}{8} \log 2 \)

Question. Evaluate: \( \int_{-1}^{3/2} |x \sin \pi x| dx \)
Answer: \( \int_{-1}^{3/2} |x \sin \pi x| dx \)
As we know
\( \sin \theta = 0 \)
\( \implies \) \( \theta = n\pi, n \in Z \)
\( \therefore \sin nx = 0 \quad x = 0, 1, 2, ... \)
For \( -1 < x < 0, x < 0, \sin \pi x < 0 \)
\( \implies \) \( x \sin \pi x > 0 \)
For \( 0 < x < 1 \)
\( x > 0, \sin \pi x > 0 \)
\( \implies \) \( x \sin \pi x > 0 \)
For \( 1 < x < \frac{3}{2}, x > 0, \sin \pi x < 0 \)
\( \implies \) \( x \sin \pi x < 0 \)
\( \therefore \int_{-1}^{3/2} |x \sin \pi x| dx = \int_{-1}^1 x \sin \pi x \, dx + \int_1^{3/2} (-x \sin \pi x) dx \)
\( = \left[ x . \frac{(-\cos \pi x)}{\pi} \right]_{-1}^1 - \int_{-1}^1 1 . \frac{-\cos \pi x}{\pi} dx - \left[ x . \frac{-\cos \pi x}{\pi} \right]_1^{3/2} + \int_1^{3/2} 1 . \frac{-\cos \pi x}{\pi} dx \)
\( = \left[ -\frac{x}{\pi} \cos \pi x + \frac{1}{\pi^2} \sin \pi x \right]_{-1}^1 - \left[ -\frac{x}{\pi} \cos \pi x + \frac{1}{\pi^2} \sin \pi x \right]_1^{3/2} \)
\( = \left[ \frac{1}{\pi} + 0 + \frac{1}{\pi} - 0 \right] - \left[ 0 - \frac{1}{\pi^2} - \frac{1}{\pi} \right] = \left[ \frac{1}{\pi} + \frac{1}{\pi} + \frac{1}{\pi^2} + \frac{1}{\pi} \right] \)
\( = \frac{1}{\pi^2} + \frac{3}{\pi} = \frac{1 + 3\pi}{\pi^2} \)

Question. Find \( \int e^x (\cos x - \sin x) \text{cosec}^2 x \, dx \).
Answer: Let \( I = \int e^x (\cos x - \sin x) \text{cosec}^2 x \, dx = \int e^x (\cot x . \text{cosec } x - \text{cosec } x) dx \)
\( = \int \underset{\text{II}}{e^x} \underset{\text{I}}{\text{cosec } x \cot x} dx - \int e^x \text{cosec } x dx \)
\( = \int e^x \text{cosec } x \cot x dx - \text{cosec } x e^x + C + \int -\text{cosec } x \cot x e^x dx \)
[Using integration by parts for 2nd integral]
\( = \int e^x \text{cosec } x \cot x dx - e^x \text{cosec } x + C - \int e^x \text{cosec } x \cot x dx \)
\( = - e^x \text{cosec } x + C. \)

Question. Evaluate: \( \int_{\pi/6}^{\pi/3} \frac{\sin x + \cos x}{\sqrt{\sin 2x}} dx \)
Answer: We have,
\( I = \int_{\pi/6}^{\pi/3} \frac{\sin x + \cos x}{\sqrt{\sin 2x}} dx \)
\( \implies \) \( I = \int_{\pi/6}^{\pi/3} \frac{\sin x + \cos x}{\sqrt{1 - (\cos x - \sin x)^2}} dx \)
Let \( t = (\cos x - \sin x) \)
\( \implies \) \( dt = - (\sin x + \cos x) dx \)
The limits are, when \( x = \frac{\pi}{6} \)
\( \implies \) \( t = \cos \frac{\pi}{6} - \sin \frac{\pi}{6} = \frac{\sqrt{3} - 1}{2} \)
and \( x = \frac{\pi}{3} \)
\( \implies \) \( t = \cos \frac{\pi}{3} - \sin \frac{\pi}{3} = \frac{1 - \sqrt{3}}{2} \)
\( \therefore I = - \int_{(\sqrt{3} - 1)/2}^{(1 - \sqrt{3})/2} \frac{1}{\sqrt{1 - t^2}} dt \)
\( = - [\sin^{-1} t]_{(\sqrt{3} - 1)/2}^{(1 - \sqrt{3})/2} = - \left[ \sin^{-1} \frac{1 - \sqrt{3}}{2} - \sin^{-1} \frac{\sqrt{3} - 1}{2} \right] = - \left[ - \sin^{-1} \frac{\sqrt{3} - 1}{2} - \sin^{-1} \frac{\sqrt{3} - 1}{2} \right] \)
\( \implies \) \( I = 2 \sin^{-1} \frac{\sqrt{3} - 1}{2} \qquad \qquad [\because \sin^{-1} (-x) = - \sin^{-1} x] \)

Question. Evaluate: \( \int_0^{\pi/4} \frac{\sin x + \cos x}{9 + 16 \sin 2x} dx \)
Answer: Let \( I = \int_0^{\pi/4} \frac{\sin x + \cos x}{9 + 16 \sin 2x} dx \)
Here, we express denominator in terms of \( \sin x - \cos x \) which is integral of the numerator.
We have, \( (\sin x - \cos x)^2 = \sin^2 x + \cos^2 x - 2 \sin x \cos x = 1 - \sin 2x \)
\( \implies \) \( \sin 2x = 1 - (\sin x - \cos x)^2 \)
\( \therefore I = \int_0^{\pi/4} \frac{\sin x + \cos x}{9 + 16(1 - (\sin x - \cos x)^2)} dx \)
\( \implies \) \( I = \int_0^{\pi/4} \frac{\sin x + \cos x}{25 - 16(\sin x - \cos x)^2} dx \)
Let \( \sin x - \cos x = t \)
\( \implies \) \( (\cos x + \sin x)dx = dt \)
The limits are, when \( x = 0 \)
\( \implies \) \( t = \sin 0 - \cos 0 = -1 \) and \( x = \frac{\pi}{4} \)
\( \implies \) \( t = \sin \frac{\pi}{4} - \cos \frac{\pi}{4} = 0 \)
\( \therefore I = \int_{-1}^0 \frac{dt}{25 - 16t^2} \)
\( \implies \) \( I = \frac{1}{16} \int_{-1}^0 \frac{dt}{\frac{25}{16} - t^2} = \frac{1}{16} \int_{-1}^0 \frac{dt}{\left(\frac{5}{4}\right)^2 - t^2} \)
\( \implies \) \( I = \frac{1}{16} \cdot \frac{1}{2\left(\frac{5}{4}\right)} \left[ \log \left| \frac{\frac{5}{4} + t}{\frac{5}{4} - t} \right| \right]_{-1}^0 \)
\( \implies \) \( I = \frac{1}{40} \left[ \log 1 - \log \left( \frac{1/4}{9/4} \right) \right] \)
\( \implies \) \( I = \frac{1}{40} \left[ 0 - \log \left( \frac{1}{9} \right) \right] = \frac{1}{40} \log 9 \)

Question. Evaluate: \( \int_0^\pi \frac{x \tan x}{\sec x + \tan x} dx \)
Answer: Let \( I = \int_0^\pi \frac{x \tan x}{\sec x + \tan x} dx \quad ... (i) \)
\( = \int_0^\pi \frac{(\pi - x)\tan(\pi - x)}{\sec(\pi - x) + \tan(\pi - x)} dx \qquad \left[ \because \int_0^a f(x)dx = \int_0^a f(a - x)dx \right] \)
\( = \int_0^\pi \frac{(\pi - x)\tan x}{\sec x + \tan x} dx \quad ... (ii) \)
By adding equations (i) and (ii), we get
\( 2I = \pi \int_0^\pi \frac{\tan x}{\sec x + \tan x} dx \)
Multiplying and dividing by \( (\sec x - \tan x) \), we get
\( 2I = \pi \int_0^\pi \frac{\tan x (\sec x - \tan x)}{\sec^2 x - \tan^2 x} dx = \pi \int_0^\pi (\sec x \tan x - \tan^2 x) dx \)
\( = \pi \int_0^\pi \sec x \tan x dx - \pi \int_0^\pi \sec^2 x dx + \int_0^\pi dx \)
\( = \pi [\sec x]_0^\pi - \pi [\tan x]_0^\pi + \pi [x]_0^\pi = \pi (-1 - 1) - 0 + \pi (\pi - 0) = \pi(\pi - 2) \)
\( \implies \) \( 2I = \pi(\pi - 2) \)
\( \implies \) \( I = \frac{\pi}{2} (\pi - 2) \)

Very Short Answer Questions

Question. Evaluate: \( \int \frac{dx}{\sqrt{1 - x^2}} \)
Answer: \( \int \frac{dx}{\sqrt{1 - x^2}} = \sin^{-1} x + C \)

Question. Write the value of \( \int \frac{dx}{x^2 + 16} \).
Answer: \( \int \frac{dx}{x^2 + 16} = \int \frac{dx}{x^2 + 4^2} \)
\( \implies \frac{1}{4} \tan^{-1} \frac{x}{4} + C \)

Question. \( \int \frac{(x^2 + 2)}{x + 1} dx \)
Answer: Let \( I = \int \frac{x^2 + 2}{x + 1} dx \) \[ = \int \left( x - 1 + \frac{3}{x + 1} \right) dx \] \[ = \int (x - 1) dx + 3 \int \frac{1}{x + 1} dx \] \[ = \frac{x^2}{2} - x + 3 \log |x + 1| + C \]

Question. Evaluate: \( \int \sec^2 (7 - x) dx \)
Answer: \( \int \sec^2 (7 - x) dx = \frac{\tan(7 - x)}{-1} + C = -\tan(7 - x) + C \)

Question. Evaluate: \( \int \frac{\sec^2 \sqrt{x}}{\sqrt{x}} dx \)
Answer: Let \( \sqrt{x} = z \implies \frac{1}{2\sqrt{x}} dx = dz \implies \frac{dx}{\sqrt{x}} = 2dz \) \[ \therefore \int \frac{\sec^2 \sqrt{x}}{\sqrt{x}} dx = 2 \int \sec^2 z dz = 2 \tan z + C = 2 \tan \sqrt{x} + C \]

Question. Evaluate: \( \int \frac{dx}{x + x \log x} \)
Answer: \( I = \int \frac{dx}{x(1 + \log x)} \) Put \( 1 + \log x = z, \implies \frac{1}{x} dx = dz \) \[ \therefore I = \int \frac{dz}{z} = \log z + C = \log (1 + \log x) + C \]

Question. If \( \int_{0}^{1} (3x^2 + 2x + k) dx = 0 \), then find the value of \( k \).
Answer: Given, \( \int_{0}^{1} (3x^2 + 2x + k) dx = 0 \)
\( \implies \left[ \frac{3x^3}{3} + \frac{2x^2}{2} + kx \right]_{0}^{1} = 0 \)
\( \implies [x^3 + x^2 + kx]_{0}^{1} = 0 \)
\( \implies (1 + 1 + k) - (0) = 0 \)
\( \implies k = -2 \)

Question. Evaluate: \( \int_{0}^{1/\sqrt{2}} \frac{dx}{\sqrt{1 - x^2}} \)
Answer: \( \int_{0}^{1/\sqrt{2}} \frac{dx}{\sqrt{1 - x^2}} = [\sin^{-1} x]_{0}^{1/\sqrt{2}} = \sin^{-1} \left( \frac{1}{\sqrt{2}} \right) - \sin^{-1}(0) = \frac{\pi}{4} - 0 = \frac{\pi}{4} \)

Question. Evaluate: \( \int_{2}^{3} \frac{1}{x} dx \)
Answer: \( \int_{2}^{3} \frac{1}{x} dx = [\log x]_{2}^{3} = \log 3 - \log 2 = \log \frac{3}{2} \)

Short Answer Questions

Question. Given \( \int e^x(\tan x + 1)\sec xdx=e^x f(x) + C \). Write \( f(x) \) satisfying the above.
Answer: Sol. Given, \( \int e^x(\tan x + 1)\sec xdx=e^x f(x) + C \)
\( \implies \) \( \int e^x (\tan x \sec x + \sec x) dx = e^x f(x) + C \)
\( \implies \) \( \int e^x (\sec x + \tan x \sec x) dx = e^x f(x) + C \)
\( \implies \) \( e^x \sec x + C = e^x f(x) + C \)
\( \implies \) \( f(x) = \sec x \) [Note: \( \int e^x [f(x) + f'(x)] dx = e^x f(x) + C \), Here \( f(x) = \sec x \)]

Question. Evaluate: \( \int (1-x)\sqrt{x} dx \)
Answer: Sol. \( \int (1-x)\sqrt{x} dx = \int \sqrt{x} dx - \int x^{1 + \frac{1}{2}} dx = \int x^{1/2} dx - \int x^{3/2} dx \)
\( = \frac{x^{1/2 + 1}}{1/2 + 1} - \frac{x^{3/2 + 1}}{3/2 + 1} + C = \frac{2}{3}x^{3/2} - \frac{2}{5}x^{5/2} + C \)

Question. Find \( \int e^x \frac{\sqrt{1 + \sin 2x}}{1 + \cos 2x} dx \).
Answer: Sol. \( I = \int e^x \frac{\sqrt{1 + \sin 2x}}{1 + \cos 2x} dx = \int e^x \frac{\sqrt{\sin^2 x + \cos^2 x + 2 \sin x \cos x}}{1 + \cos 2x} dx \)
\( = \int e^x \frac{\sqrt{(\sin x + \cos x)^2}}{1 + \cos 2x} dx = \int e^x \frac{\sin x + \cos x}{2 \cos^2 x} dx = \frac{1}{2} \int e^x \left( \frac{\sin x}{\cos^2 x} + \frac{\cos x}{\cos^2 x} \right) dx \)
\( = \frac{1}{2} \int e^x (\sec x \tan x + \sec x) dx \)
\( = \frac{1}{2} \int e^x (\sec x + \sec x \cdot \tan x) dx = \frac{1}{2} e^x \cdot \sec x + C \) [\( \because \int e^x (f(x) + f'(x)) dx = e^x f(x) + C \)]

Question. Find \( \int \frac{x-1}{(x-2)(x-3)} dx \).
Answer: Sol. \( \because \frac{x-1}{(x-2)(x-3)} = \frac{A}{x-2} + \frac{B}{x-3} \) where \( A = \left. \frac{x-1}{x-3} \right|_{x=2} = \frac{1}{-1} = -1 \) & \( B = \left. \frac{x-1}{x-2} \right|_{x=3} = \frac{2}{1} = 2 \)
\( \implies \) \( \frac{x-1}{(x-2)(x-3)} = \frac{-1}{(x-2)} + \frac{2}{(x-3)} \)
\( \implies \) \( \int \frac{x-1}{(x-2)(x-3)} dx = -\int \frac{dx}{x-2} + 2 \int \frac{dx}{x-3} = -\log (x-2) + 2 \log (x-3) + C \)
\( \implies \) \( \int \frac{x-1}{(x-2)(x-3)} dx = -\log (x-2) + \log (x-3)^2 + C = \log \frac{(x-3)^2}{(x-2)} + C \)

Question. Find \( \int_{-\pi/4}^0 \frac{1 + \tan x}{1 - \tan x} dx \).
Answer: Sol. \( \int_{-\pi/4}^0 \frac{1 + \tan x}{1 - \tan x} dx = \int_{-\pi/4}^0 \tan \left( \frac{\pi}{4} + x \right) dx \)
\( = \left[ \log \sec \left( \frac{\pi}{4} + x \right) \right]_{-\pi/4}^0 = \log \sec \left( \frac{\pi}{4} \right) - \log \sec \left( \frac{\pi}{4} - \frac{\pi}{4} \right) \)
\( = \log (\sqrt{2}) - \log (\sec (0)) = \log (\sqrt{2}) - \log 1 \)
\( = \log \sqrt{2} = \frac{1}{2} \log 2 \)

Question. Find \( \int \frac{dx}{\sqrt{5-4x-2x^2}} \).
Answer: Sol. \( \int \frac{dx}{\sqrt{5-4x-2x^2}} = \int \frac{dx}{\sqrt{7 - 2 - 4x - 2x^2}} = \int \frac{dx}{\sqrt{7 - 2(1 + 2x + x^2)}} = \frac{1}{\sqrt{2}} \int \frac{dx}{\sqrt{\frac{7}{2} - (x+1)^2}} \)
\( = \frac{1}{\sqrt{2}} \int \frac{dx}{\sqrt{\left( \sqrt{\frac{7}{2}} \right)^2 - (x+1)^2}} = \frac{1}{\sqrt{2}} \sin^{-1} \left( \frac{x+1}{\sqrt{7/2}} \right) + C = \frac{1}{\sqrt{2}} \sin^{-1} \left( \sqrt{\frac{2}{7}} (x+1) \right) + C \)

Question. Evaluate \( \int_0^1 \frac{\tan^{-1} x}{1 + x^2} dx \).
Answer: Sol. Let \( t = \tan^{-1} x \implies dt = \frac{1}{1 + x^2} dx \) Also when, \( x = 0, t = 0 \) and when \( x = 1 \),
\( \implies \) \( t = \frac{\pi}{4} \)
\( \therefore \int_0^1 \frac{\tan^{-1} x}{1 + x^2} dx = \int_0^{\pi/4} t dt = \left[ \frac{t^2}{2} \right]_0^{\pi/4} = \frac{1}{2} \left[ \frac{\pi^2}{16} - 0 \right] = \frac{\pi^2}{32} \)

Question. Evaluate: \( \int_0^1 \frac{dx}{\sqrt{2x+3}} \)
Answer: Sol. Let \( I = \int_0^1 \frac{dx}{\sqrt{2x+3}} = \int_0^1 (2x+3)^{-1/2} dx = \left[ \frac{(2x+3)^{-1/2+1}}{(-1/2+1) \times 2} \right]_0^1 = \left[ \frac{(2x+3)^{1/2}}{\frac{1}{2} \times 2} \right]_0^1 = 5^{1/2} - 3^{1/2} = \sqrt{5} - \sqrt{3} \)

Long Answer Questions-I

Question. Evaluate: \( \int \frac{2x}{(x^2 + 1)(x^2 + 3)} dx \)
Answer: Sol. Let \( x^2 = z \implies 2x dx = dz \)
\( \therefore \int \frac{2x}{(x^2 + 1)(x^2 + 3)} dx = \int \frac{dz}{(z+1)(z+3)} \) Using partial fraction. Let \( \frac{1}{(z+1)(z+3)} = \frac{A}{z+1} + \frac{B}{z+3} \) ...(i)
\( \implies \) \( \frac{1}{(z+1)(z+3)} = \frac{A(z+3) + B(z+1)}{(z+1)(z+3)} \)
\( \implies \) \( 1 = A(z+3) + B(z+1) \implies 1 = (A+B)z + (3A+B) \) Equating the coefficient of z and constant, we get \( A + B = 0 \) ...(ii) and \( 3A + B = 1 \) ...(iii) Subtracting (ii) from (iii), we get \( 2A = 1 \implies A = \frac{1}{2} \) and \( B = -\frac{1}{2} \) Putting the values of A and B in (i), we get \( \frac{1}{(z+1)(z+3)} = \frac{1}{2(z+1)} - \frac{1}{2(z+3)} \)
\( \therefore \int \frac{2x dx}{(x^2+1)(x^2+3)} = \int \left( \frac{1}{2(z+1)} - \frac{1}{2(z+3)} \right) dz = \frac{1}{2} \int \frac{dz}{z+1} - \frac{1}{2} \int \frac{dz}{z+3} \)
\( = \frac{1}{2} \log |z+1| - \frac{1}{2} \log |z+3| + C = \frac{1}{2} \log |x^2+1| - \frac{1}{2} \log |x^2+3| + C \)
\( = \frac{1}{2} \log \left| \frac{x^2+1}{x^2+3} \right| + C = \log \sqrt{\frac{x^2+1}{x^2+3}} + C \) [Note: \( \log m + \log n = \log m \cdot n \) and \( \log m - \log n = \log m / n \)]

Question. Evaluate: \( \int \frac{x^2}{1-x^4} dx \)
Answer: Sol. Let \( I = \int \frac{x^2}{1-x^4} dx = \int \frac{1}{2} \frac{x^2+1+x^2-1}{(1-x^2)(1+x^2)} dx \) [\( \because a^2 - b^2 = (a+b)(a-b) \)]
\( = \int \frac{\frac{1}{2}(1+x^2) - \frac{1}{2}(1-x^2)}{(1-x^2)(1+x^2)} dx = \int \frac{\frac{1}{2}(1+x^2)}{(1-x^2)(1+x^2)} dx - \int \frac{\frac{1}{2}(1-x^2)}{(1-x^2)(1+x^2)} dx \)
\( = \frac{1}{2} \int \frac{1}{1-x^2} dx - \frac{1}{2} \int \frac{1}{1+x^2} dx = \frac{1}{2} \cdot \frac{1}{2} \log \left| \frac{1+x}{1-x} \right| + C_1 - \frac{1}{2} \tan^{-1} x + C_2 \)
\( = \frac{1}{4} \log \left| \frac{1+x}{1-x} \right| - \frac{1}{2} \tan^{-1} x + C \) [\( \because C = C_1 + C_2 \)]