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

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  $\tan 2\theta$

Solution:

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

Let  $\mathrm{x}=\tan \theta$

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

Now, $0 < x <1$

$0<\tan \theta<1$

$0<\theta<\frac{\pi}{4}$

Now,

\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}

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

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

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