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  • Need solution for RD Sharma Maths Class 12 Chapter Definite Integrals exercise 19.2 question 17.

Need solution for RD Sharma Maths Class 12 Chapter Definite Integrals exercise 19.2 question 17.

Answers (1)

Answer:\frac{\pi}{2}-\log 2

Hint: We use indefinite integral formula then put limits to solve this integral.

Given: \int_{0}^{1}\tan ^{-1}\left ( \frac{2x}{1-x^2} \right )dx

Solution: I=\int_{0}^{1}\tan ^{-1}\left ( \frac{2x}{1-x^2} \right )dx

Put x=\tan \theta

dx=\sec^2\theta \; d\theta

When x=0  then \theta=0  and

when x=1  then \theta=\frac{\pi}{4}

\begin{aligned} &I=\int_{0}^{\frac{\pi}{4}} \tan ^{-1}\left(\frac{2 \tan \theta}{1-\tan ^{2} \theta}\right) \sec ^{2} \theta d \theta\; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\tan 2 \theta=\frac{2 \tan \theta}{1-\tan ^{2} \theta}\right] \\ &=\int_{0}^{\frac{\pi}{4}} \tan ^{-1}(\tan 2 \theta) \sec ^{2} \theta d \theta \\ &=\int_{0}^{\frac{\pi}{4}} 2 \theta \sec ^{2} \theta d \theta \\ &=2 \int_{0}^{\frac{\pi}{4}} \theta \sec ^{2} \theta d \theta \end{aligned}

Applying integration by parts method, then

\begin{aligned} &=2\left\{\left[\theta \int \sec ^{2} \theta d \theta\right]_{0}^{\frac{\pi}{4}}-\int_{0}^{\frac{\pi}{4}}\left[\frac{d(\theta)}{d \theta} \int \sec ^{2} \theta d \theta\right] d \theta\right\} \\ &=2\left\{[\theta \tan \theta]_{0}^{\frac{\pi}{4}}-\int_{0}^{\frac{\pi}{4}} 1 \tan \theta d \theta\right\} \quad\left[\int \sec ^{2} x d x=\tan x, \frac{d\left(x^{n}\right)}{d x}=n x^{n-1}\right] \\ &=2\left\{\left[\frac{\pi}{4} \tan \frac{\pi}{4}-0 \times \tan 0\right]-\int_{0}^{\frac{\pi}{4}} \tan \theta d \theta\right\} \quad\left[\int \tan x d x=-\log |\cos x|, \tan \frac{\pi}{4}=1, \tan 0=0\right] \end{aligned}

\begin{aligned} &=2\left\{\left[\frac{\pi}{4} \cdot 1-0\right]-[-\log |\cos \theta|]_{0}^{\frac{\pi}{4}}\right\} \\ &=\frac{2 \pi}{4}+2\left[\log \frac{1}{\sqrt{2}}-\log 1\right] \; \; \; \; \; \; \; \: \: \: \: \quad\left[\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}, \cos 0=1\right] \\ &=\frac{\pi}{2}+2\left[\log \frac{1}{\sqrt{2}}-0\right] \: \: \: \: \: \: \: \; \; \; \; \; \; \; \; \; \; \; \; \; \quad[\log 1=0] \end{aligned}

\begin{aligned} &=\frac{\pi}{2}+2 \log (2)^{-\frac{1}{2}} \\ &=\frac{\pi}{2}+\left(-\frac{1}{2}\right) \times 2 \log 2 \quad\left[\log a^{m}=m \log a\right] \\ &=\frac{\pi}{2}-\log 2 \end{aligned}

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