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Need solution for RD Sharma Maths Class 12 Chapter 19 Definite Integrals Excercise Revision Exercise Question 7

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Answer:  \frac{\pi}{2}-\log 2

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

Hint: Use substitution method and then apply the formula of  \tan 2\theta


\int_{0}^{1} \tan ^{-1}\left(\frac{2 x}{1+x^{2}}\right) d x

Let  \mathrm{x}=\tan \theta

d x=\sec ^{2} \theta d \theta                     (differentiate w.r.t x )

Now, 0 < x <1

0<\tan \theta<1



\begin{aligned} &\int_{0}^{\frac{\pi}{4}} \tan ^{-1}\left(\frac{2 \tan \theta}{1+\tan ^{2} \theta}\right) \sec ^{2} \theta d \theta \\ & \end{aligned}

=2 \int_{0}^{\frac{\pi}{4}} \theta \sec ^{2} \theta d \theta \\

=2\left[\theta \int \sec ^{2} \theta d \theta-\int \frac{d}{d \theta} \theta \int \sec ^{2} \theta d \theta\right]_{0}^{\frac{\pi}{4}}

\begin{aligned} &{\left[\because \int u v d x=u \int v d x-\int \frac{d}{d x} u \int v d x\right]} \\ & \end{aligned}

=2\left[(\theta \tan \theta)_{0}^{\frac{\pi}{4}}-\int_{0}^{\frac{\pi}{4}} \tan \theta d \theta\right]

\begin{aligned} &=2\left(\frac{\pi}{4}-0\right)-2 \int_{0}^{\frac{\pi}{4}} \tan \theta d \theta \\ & \end{aligned}

=\frac{\pi}{2}-2 \int_{0}^{\frac{\pi}{4}} \frac{\sin \theta}{\cos \theta} d \theta

Let  \cos \theta=t

-\sin \theta d \theta=\mathrm{dt}                           (differentiate w.r.t )

\begin{aligned} &=\frac{\pi}{2}+2 \int_{0}^{\frac{\pi}{4}} \frac{1}{t} d t \\ & \end{aligned}

=\frac{\pi}{2}+2(\log |t|)_{0}^{\frac{\pi}{4}}

\begin{aligned} &=\frac{\pi}{2}+2(\log |\cos \theta|)_{0}^{\frac{\pi}{4}}\\ & \end{aligned}

=\frac{\pi}{2}+2\left(\log \frac{1}{\sqrt{2}}-\log 1\right)\\

=\frac{\pi}{2}+2 \log \frac{1}{\sqrt{2}}\\                            (\because \log 1=0)\\

=\frac{\pi}{2}-\log 2

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