CBSE Class 12 Mathematics Indefinite And Definite Integrals Worksheet Set 02

CBSE Class 12 Mathematics Indefinite and Definite Integrals (2). Students can download these worksheets and practice them. This will help them to get better marks in examinations. Also refer to other worksheets for the same chapter and other subjects too. Use them for better understanding of the subjects.

Question. \[ \int_{0}^{1} \tan^{-1} \left( \frac{2x-1}{1+x-x^2} \right) \, dx \]
Answer:
\[ I = \int_{0}^{1} \tan^{-1} \left( \frac{2x-1}{1+x-x^2} \right) \, dx \quad \dots (1) \]
\[ I = \int_{0}^{1} \tan^{-1} \left( \frac{2(1-x)-1}{1+(1-x)-(1-x)^2} \right) \, dx \quad \dots (\text{P-IV}) \]
\[ I = \int_{0}^{1} \tan^{-1} \left( \frac{2-2x-1}{1+1-x-1-x^2+2x} \right) \, dx \]
\[ I = \int_{0}^{1} \tan^{-1} \left( \frac{-2x+1}{1+x-x^2} \right) \, dx \]
\[ I = - \int_{0}^{1} \tan^{-1} \left( \frac{2x-1}{1+x-x^2} \right) \, dx \quad \dots (2) \quad \dots \{\because \tan^{-1}(-x) = -\tan^{-1}x\} \]
Adding (1) and (2):
\[ 2I = 0 \]
\[ I = 0 \quad \text{ans.} \]

Alternate:
\[ I = \int_{0}^{1} \tan^{-1} \left( \frac{2x-1}{1+x-x^2} \right) \, dx \]
\[ I = \int_{0}^{1} \tan^{-1} \left( \frac{(x)+(x-1)}{1-x(x-1)} \right) \, dx \quad \dots \{\text{adjustment}\} \]
\[ I = \int_{0}^{1} \tan^{-1}x \, dx + \int_{0}^{1} \tan^{-1}(x-1) \, dx \quad \dots \left[\tan^{-1}\left(\frac{x+y}{1-xy}\right) = \tan^{-1}x + \tan^{-1}y\right] \]
\[ I = \int_{0}^{1} \tan^{-1}x \, dx + \int_{0}^{1} \tan^{-1}[(1-x)-1] \, dx \quad \dots (\text{P-IV}) \]
\[ I = \int_{0}^{1} \tan^{-1}x \, dx + \int_{0}^{1} \tan^{-1}(-x) \, dx \]
\[ I = \int_{0}^{1} \tan^{-1}x \, dx - \int_{0}^{1} \tan^{-1}x \, dx \quad \dots [\tan^{-1}(-x) = -\tan^{-1}x] \]
\[ I = 0 \quad \text{ans.} \]

Question. \[ \int_{0}^{1} \cot^{-1}(1 - x + x^2) \, dx \]
Answer:
\[ I = \int_{0}^{1} \cot^{-1}(1 - x + x^2) \, dx \]
\[ I = \int_{0}^{1} \tan^{-1} \left( \frac{1}{1-x+x^2} \right) \, dx \quad \dots \left[ \tan^{-1}\left(\frac{1}{x}\right) = \cot^{-1}x \right] \]
\[ I = \int_{0}^{1} \tan^{-1} \left[ \frac{x+(1-x)}{1-x(1-x)} \right] \, dx \quad \dots [\text{adjustment}] \]
\[ I = \int_{0}^{1} \tan^{-1}(x) \, dx + \int_{0}^{1} \tan^{-1}(1-x) \, dx \quad \dots \left[ \tan^{-1}\left(\frac{x+y}{1-xy}\right) = \tan^{-1}x + \tan^{-1}y \right] \]
\[ I = \int_{0}^{1} \tan^{-1}(x) \, dx + \int_{0}^{1} \tan^{-1}[1-(1-x)] \, dx \quad \dots (\text{P-IV}) \]
\[ I = \int_{0}^{1} \tan^{-1}x \, dx + \int_{0}^{1} \tan^{-1}(x) \, dx \]
\[ I = 2 \int_{0}^{1} \tan^{-1}x \, dx \]
\[ I = 2 \int_{0}^{1} \tan^{-1}x \cdot 1 \, dx \]
\[ I = 2 \left[ (x \tan^{-1}x)_{0}^{1} - \int_{0}^{1} \frac{1}{1+x^2} \cdot x \, dx \right] \]
\[ I = 2 \left[ \left(\frac{\pi}{4} - 0\right) - \int_{0}^{1} \frac{x}{1+x^2} \, dx \right] \]
Put \( 1 + x^2 = t \) when \( x = 0 \Rightarrow t = 1 \)
\( x \, dx = \frac{dt}{2} \) when \( x = 1 \Rightarrow t = 2 \)
\[ \therefore I = 2 \left[ \frac{\pi}{4} - \frac{1}{2} \int_{1}^{2} \frac{dt}{t} \right] \]
\[ = \frac{\pi}{2} - \int_{1}^{2} \frac{dt}{t} \]
\[ = \frac{\pi}{2} - [\log t]_{1}^{2} \]
\[ = \frac{\pi}{2} - [\log 2 - \log 1] \]
\[ I = \frac{\pi}{2} - \log 2 \quad \text{ans.} \quad [\because \log(1) = 0] \]

Question. \[ \int_{0}^{\pi} \frac{x \sin x}{1+\cos^2x} \, dx \quad \dots [\text{Removal of } x] \]
Answer:
\[ I = \int_{0}^{\pi} \frac{x \sin x}{1+\cos^2x} \, dx \quad \dots (1) \]
\[ I = \int_{0}^{\pi} \frac{(\pi-x)\sin(\pi-x)}{1+\cos^2(\pi-x)} \, dx \quad \dots (\text{P-IV}) \]
\[ I = \int_{0}^{\pi} \frac{(\pi-x)\sin x}{1+\cos^2x} \, dx \quad \dots (2) \quad \left\{ \begin{aligned} \cos(\pi-x) &= -\cos x \\ \sin(\pi-x) &= \sin x \end{aligned} \right\} \]
Adding (1) and (2):
\[ 2I = \int_{0}^{\pi} \frac{x \sin x + \pi \sin x - x \sin x}{1+\cos^2x} \, dx \]
\[ 2I = \pi \int_{0}^{\pi} \frac{\sin x}{1+\cos^2x} \, dx \]
Put \( \cos x = t \) when \( x = 0 \Rightarrow t = 1 \)
\( \sin x \, dx = -dt \) when \( x = \pi \Rightarrow t = -1 \)
\[ \therefore 2I = -\pi \int_{1}^{-1} \frac{dt}{1+t^2} \]
\[ 2I = -\pi [\tan^{-1} t]_{1}^{-1} \]
\[ 2I = -\pi [\tan^{-1}(-1) - \tan^{-1}(1)] \]
\[ 2I = -\pi \left[ -\frac{\pi}{4} - \frac{\pi}{4} \right] \]
\[ 2I = -\pi \left( -\frac{\pi}{2} \right) \]
\[ \therefore I = \frac{\pi^2}{4} \quad \text{ans.} \]

Question. \[ \int_{0}^{\pi} \frac{x \tan x}{\sec x+\tan x} \, dx \]
Answer:
\[ I = \int_{0}^{\pi} \frac{x \tan x}{\sec x + \tan x} \, dx \quad \dots \{\text{change in } \sin x \text{ \& } \cos x\} \]
\[ I = \int_{0}^{\pi} \frac{x \sin x}{1 + \sin x} \, dx \quad \dots (1) \]
\[ I = \int_{0}^{\pi} \frac{(\pi-x)\sin(\pi-x)}{1+\sin(\pi-x)} \, dx \quad \dots (\text{P-IV}) \]
\[ I = \int_{0}^{\pi} \frac{(\pi-x)\sin x}{1+\sin x} \, dx \quad \dots (2) \]
Adding (1) and (2):
\[ 2I = \int_{0}^{\pi} \frac{x \sin x + \pi \sin x - x \sin x}{1+\sin x} \, dx \]
\[ 2I = \pi \int_{0}^{\pi} \frac{\sin x}{1+\sin x} \, dx \]
Type: rationalize
\[ 2I = \pi \int_{0}^{\pi} \frac{\sin x}{1+\sin x} \times \frac{(1-\sin x)}{(1-\sin x)} \, dx \]
\[ 2I = \pi \int_{0}^{\pi} \frac{\sin x - \sin^2x}{\cos^2x} \, dx \]
\[ 2I = \pi \int_{0}^{\pi} \left( \frac{\sin x}{\cos^2x} - \frac{\sin^2x}{\cos^2x} \right) \, dx \]
\[ 2I = \pi \int_{0}^{\pi} (\tan x \sec x - \tan^2x) \, dx \]
\[ 2I = \pi \int_{0}^{\pi} [\tan x \sec x - (\sec^2x - 1)] \, dx \]
\[ 2I = \pi [\sec x - \tan x + x]_{0}^{\pi} \]
\[ 2I = \pi [(\sec \pi - \tan \pi + \pi) - (\sec 0 - \tan 0 + 0)] \]
\[ 2I = \pi [(-1 - 0 + \pi) - (1 - 0)] \]
\[ 2I = \pi [\pi - 2] \]
\[ \therefore I = \frac{\pi}{2} (\pi - 2) \quad \text{ans.} \]

Question. \[ \int_{0}^{\frac{\pi}{2}} \frac{x \sin x \cos x}{\sin^4x+\cos^4x} \, dx \]
Answer:
\[ I = \int_{0}^{\frac{\pi}{2}} \frac{x \sin x \cdot \cos x}{\sin^4x+\cos^4x} \, dx \quad \dots (1) \]
\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\left(\frac{\pi}{2}-x\right)\sin\left(\frac{\pi}{2}-x\right)\cdot\cos\left(\frac{\pi}{2}-x\right)}{\sin^4\left(\frac{\pi}{2}-x\right)+\cos^4\left(\frac{\pi}{2}-x\right)} \, dx \quad \dots (\text{P-IV}) \]
\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\left(\frac{\pi}{2}-x\right)\cos x \cdot \sin x}{\cos^4x+\sin^4x} \, dx \quad \dots (2) \]
Adding (1) and (2):
\[ 2I = \int_{0}^{\frac{\pi}{2}} \frac{x \sin x \cos x + \left(\frac{\pi}{2}-x\right)\sin x \cos x}{\sin^4x+\cos^4x} \, dx \]
\[ 2I = \frac{\pi}{2} \int_{0}^{\frac{\pi}{2}} \frac{\sin x \cdot \cos x}{\sin^4x+\cos^4x} \, dx \]
Divide N & D by \( \cos^4 x \):
\[ 2I = \frac{\pi}{2} \int_{0}^{\frac{\pi}{2}} \frac{\tan x \sec^2x}{\tan^4x+1} \, dx \]
Put \( \tan^2x = t \) when \( x = 0 \Rightarrow t = 0 \)
\( 2 \tan x \sec^2x \, dx = dt \) when \( x = \frac{\pi}{2} \Rightarrow t = \infty \)
\( \tan x \sec^2x \, dx = \frac{dt}{2} \)
\[ \therefore 2I = \frac{\pi}{4} \int_{0}^{\infty} \frac{dt}{t^2 + 1} \]
\[ 2I = \frac{\pi}{4} [\tan^{-1}t]_{0}^{\infty} \]
\[ 2I = \frac{\pi}{4} [\tan^{-1}(\infty) - \tan^{-1}(0)] \]
\[ 2I = \frac{\pi}{4} \left[ \frac{\pi}{2} - 0 \right] \]
\[ \Rightarrow 2I = \frac{\pi^2}{8} \]
\[ \Rightarrow I = \frac{\pi^2}{16} \quad \text{ans.} \]

Question. \[ I = \int_{0}^{\pi} \frac{x \, dx}{a^2\cos^2x + b^2\sin^2x} \]
Answer:
\[ I = \int_{0}^{\pi} \frac{x \, dx}{a^2\cos^2x + b^2\sin^2x} \quad \dots (1) \]
\[ I = \int_{0}^{\pi} \frac{(\pi-x) \, dx}{a^2\cos^2(\pi-x) + b^2\sin^2(\pi-x)} \quad \dots (\text{P-IV}) \]
\[ I = \int_{0}^{\pi} \frac{(\pi-x) \, dx}{a^2\cos^2x + b^2\sin^2x} \quad \dots (2) \]
Adding (1) and (2):
\[ 2I = \int_{0}^{\pi} \frac{x + \pi - x}{a^2\cos^2x + b^2\sin^2x} \, dx \]
\[ 2I = \pi \int_{0}^{\pi} \frac{1}{a^2\cos^2x + b^2\sin^2x} \, dx \]
Type: Divide by \( \cos^2 x \)
Divide N & D by \( \cos^2x \):
\[ \therefore 2I = \pi \int_{0}^{\pi} \frac{\sec^2x}{a^2 + b^2\tan^2x} \, dx \]
\[ 2I = 2\pi \int_{0}^{\frac{\pi}{2}} \frac{\sec^2x}{a^2 + b^2\tan^2x} \, dx \quad \dots \left[ \int_{0}^{2a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx \right] \quad \dots (\text{P-VI}) \]
Put \( \tan x = t \) when \( x = 0 \Rightarrow t = 0 \)
\( \sec^2x \, dx = dt \) when \( x = \frac{\pi}{2} \Rightarrow t = \infty \)
\[ \dots 2I = 2\pi \int_{0}^{\infty} \frac{dt}{a^2 + b^2t^2} \]
\[ I = \frac{\pi}{b^2} \int_{0}^{\infty} \frac{dt}{\left(\frac{a}{b}\right)^2 + t^2} \]
\[ I = \frac{\pi}{b^2} \times \frac{b}{a} \left[ \tan^{-1}\left(\frac{bt}{a}\right) \right]_{0}^{\infty} \]
\[ I = \frac{\pi}{ab} [\tan^{-1}(\infty) - \tan^{-1}(0)] \]
\[ I = \frac{\pi}{ab} \left[ \frac{\pi}{2} - 0 \right] \]
\[ I = \frac{\pi^2}{2ab} \quad \text{ans.} \]

Question. \[ I = \int_{0}^{\pi} \frac{x}{1-\cos\alpha \sin x} \, dx \]
Answer:
\[ I = \int_{0}^{\pi} \frac{x}{1-\cos\alpha \sin x} \, dx \quad \dots (1) \]
\[ I = \int_{0}^{\pi} \frac{\pi - x}{1-\cos\alpha \sin(\pi-x)} \, dx \quad \dots (\text{P-IV}) \]
\[ I = \int_{0}^{\pi} \frac{\pi - x}{1-\cos\alpha \sin x} \, dx \quad \dots (2) \]
Adding (1) and (2):
\[ 2I = \int_{0}^{\pi} \frac{x + \pi - x}{1-\cos\alpha \sin x} \, dx \]
\[ 2I = \pi \int_{0}^{\pi} \frac{1}{1-\cos\alpha \sin x} \, dx \]
(Type: single \( \sin x \), \( \cos x \))
\[ 2I = \pi \int_{0}^{\pi} \frac{1}{1-\cos\alpha \cdot \frac{2\tan\frac{x}{2}}{1+\tan^2\frac{x}{2}}} \, dx \]
\[ 2I = \pi \int_{0}^{\pi} \frac{1+\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2} - 2\cos\alpha \cdot \tan\frac{x}{2}} \, dx \]
\[ 2I = \pi \int_{0}^{\pi} \frac{\sec^2\frac{x}{2}}{1+\tan^2\frac{x}{2}-2\cos\alpha \cdot \tan\frac{x}{2}} \, dx \]
Put \( \tan\frac{x}{2} = t \) when \( x = 0 \Rightarrow t = 0 \)
\( \sec^2\frac{x}{2} \, dx = 2\,dt \) when \( x = \pi \Rightarrow t = \infty \)
\[ 2I = \frac{\pi}{2} \int_{0}^{\infty} \frac{dt}{t^2 - \frac{2t}{\cos\alpha} + 1} \]
\[ 2I = \frac{\pi}{2} \int_{0}^{\infty} \frac{dt}{t^2-2\cos\alpha \cdot t+1} \]
(perfect square)
\[ 2I = \frac{\pi}{2} \int_{0}^{\infty} \frac{1}{(t-\cos\alpha)^2-\cos^2\alpha+1} \, dt \]
\[ 2I = \frac{\pi}{2} \int_{0}^{\infty} \frac{dt}{(t-\cos\alpha)^2-\sin^2\alpha} \quad \dots [1 - \cos^2\alpha = \sin^2\alpha] \]
\[ 2I = \frac{\pi}{2} \times \frac{1}{\sin\alpha} \left[ \tan^{-1}\left(\frac{t-\cos\alpha}{\sin\alpha}\right) \right]_{0}^{\infty} \]
\[ 2I = \frac{\pi}{2\sin\alpha} \left[ \tan^{-1}(\infty) - \tan^{-1}\left(\frac{-\cos\alpha}{\sin\alpha}\right) \right] \]
\[ I = \frac{\pi}{4\sin\alpha} \left[ \frac{\pi}{2} - \tan^{-1}(-\cot\alpha) \right] \]
\[ I = \frac{\pi}{4\sin\alpha} \left[ \frac{\pi}{2} + \tan^{-1}(\cot\alpha) \right] \quad \dots [\tan^{-1}(-x) = -\tan^{-1}x] \]
\[ I = \frac{\pi}{4\sin\alpha} \left[ \frac{\pi}{2} + \tan^{-1}\left(\tan\left(\frac{\pi}{2}-\alpha\right)\right) \right] \]
\[ I = \frac{\pi}{4\sin\alpha} \left[ \frac{\pi}{2} + \frac{\pi}{2} - \alpha \right] \]
\[ I = \frac{\pi}{4\sin\alpha} [\pi - \alpha] \quad \text{ans.} \]

Question. \[ I = \int_{0}^{\infty} \frac{\log x}{1+x^2} \, dx \]
Answer:
Put \( x = \tan\theta \) when \( x = 0 \Rightarrow \theta = 0 \)
\( dx = \sec^2\theta \, d\theta \) when \( x = \infty \Rightarrow \theta = \frac{\pi}{2} \)
\[ \therefore I = \int_{0}^{\frac{\pi}{2}} \frac{\log(\tan\theta)}{1+\tan^2\theta} \cdot \sec^2\theta \, d\theta \]
\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\log(\tan\theta)}{\sec^2\theta} \cdot \sec^2\theta \, d\theta \]
\[ I = \int_{0}^{\frac{\pi}{2}} \log(\tan\theta) \, d\theta \quad \dots (1) \]
Proceed Yourself
\( 0 \quad \text{ans.} \]

Question. \[ I = \int_{0}^{1} \frac{\log(1+x)}{1+x^2} \, dx \]
Answer:
Put \( x = \tan\theta \) when \( x = 0 \Rightarrow \theta = 0 \)
\( dx = \sec^2\theta \, d\theta \) when \( x = 1 \Rightarrow \theta = \frac{\pi}{4} \)
\[ \therefore I = \int_{0}^{\frac{\pi}{4}} \frac{\log(1+\tan\theta)}{1+\tan^2\theta} \cdot \sec^2\theta \, d\theta \]
\[ I = \int_{0}^{\frac{\pi}{4}} \log(1+\tan\theta) \, d\theta \quad \dots (1) \]
Proceed yourself
\[ I = \frac{\pi}{8} \log 2 \quad \text{ans.} \]

Question. \[ I = \int_{0}^{\frac{\pi}{2}} \log(\sin x) \, dx \]
Answer:
\[ I = \int_{0}^{\frac{\pi}{2}} \log(\sin x) \, dx \quad \dots (1) \]
\[ I = \int_{0}^{\frac{\pi}{2}} \log\left(\sin\left(\frac{\pi}{2}-x\right)\right) \, dx \quad \dots (\text{P-IV}) \]
\[ I = \int_{0}^{\frac{\pi}{2}} \log(\cos x) \, dx \quad \dots (2) \]
Adding (1) and (2):
\[ 2I = \int_{0}^{\frac{\pi}{2}} \log(\sin x \cdot \cos x) \, dx \]
\[ 2I = \int_{0}^{\frac{\pi}{2}} \log\left(\frac{\sin(2x)}{2}\right) \, dx \]
\[ 2I = \int_{0}^{\frac{\pi}{2}} \log(\sin(2x)) - \log 2 \, dx \]
\[ 2I = \int_{0}^{\frac{\pi}{2}} \log(\sin(2x)) \, dx - \int_{0}^{\frac{\pi}{2}} \log 2 \cdot dx \]
\[ 2I = \int_{0}^{\frac{\pi}{2}} \log(\sin(2x)) \, dx - \log 2 (x)_{0}^{\frac{\pi}{2}} \]
\[ 2I = \int_{0}^{\frac{\pi}{2}} \log(\sin(2x)) \, dx - \frac{\pi}{2} \log 2 \]
\[ 2I = I_1 - \frac{\pi}{2}\log 2 \quad \dots (3) \]
Where \( I_1 = \int_{0}^{\frac{\pi}{2}} \log(\sin(2x)) \, dx \)
Put \( 2x = t \) when \( x = 0 \Rightarrow t = 0 \)
\( dx = \frac{dt}{2} \quad x = \frac{\pi}{2} \Rightarrow t = \pi \)
\[ \dots I_1 = \frac{1}{2} \int_{0}^{\pi} \log(\sin t) \, dt \]
\[ I_1 = \frac{1}{2} \times 2 \int_{0}^{\frac{\pi}{2}} \log(\sin t) \, dt \quad \dots (\text{P-VI}) \]
\[ I_1 = \int_{0}^{\frac{\pi}{2}} \log(\sin t) \, dt \]
\[ I_1 = \int_{0}^{\frac{\pi}{2}} \log(\sin x) \, dx \quad \dots (\text{P-I}) \]
\[ I_1 = I \]
\( \therefore \) eq. (3) becomes:
\[ 2I = I - \frac{\pi}{2} \log 2 \]
\[ \therefore I = -\frac{\pi}{2} \log 2 \quad \text{ans.} \]