#### Need solution for RD Sharma Maths Class 12 Chapter 19 Definite Integrals Excercise Revision Exercise Question 24

Answer:  $\frac{\pi^{2}}{16}-\frac{\pi}{4}+\frac{1}{2} \log 2$

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

Hint:  Apply the formula of  $\int uv dx$

Solution:

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

Let

\begin{aligned} &\tan ^{-1} x=t \\ & \end{aligned}

$x=\tan t \\$                                        (Differentiate w.r.t to x)

$d x=\sec ^{2} t d t$

\begin{aligned} &=\int_{0}^{1}(\tan t) t^{2} \sec ^{2} t d t \\ & \end{aligned}

$=\left[t^{2} \frac{\tan ^{2} t}{2}\right]_{0}^{1}-\int_{0}^{1} 2 t \times \frac{\tan ^{2} t}{2} d t \\$

$=\left[t^{2} \frac{\tan ^{2} t}{2}\right]_{0}^{1}-\int_{0}^{1} t\left(\sec ^{2} t-1\right) d t$

\begin{aligned} &=\left[t^{2} \frac{\tan ^{2} t}{2}\right]_{0}^{1}-\int_{\ell}^{1} t \sec ^{2} t d t+\int_{0}^{1} t d t \\ & \end{aligned}

$=\left[t^{2} \frac{\tan ^{2} t}{2}\right]_{0}^{1}-(t \tan t)_{0}^{1}+\int_{0}^{1} \tan t d t+\left(\frac{t^{2}}{2}\right)_{0}^{1} \\$

$=\left[t^{2} \frac{\tan ^{2} t}{2}\right]_{0}^{1}-(t \tan t)_{0}^{1}+[\log |\sec t|]_{0}^{1}+\left(\frac{t^{2}}{2}\right)_{0}^{1}$

Now  $0< x< 1$

\begin{aligned} &\Rightarrow 0<\tan ^{-1} x<\frac{\pi}{4} \\ & \end{aligned}

$\Rightarrow 0

\begin{aligned} &=\left[t^{2} \frac{\tan ^{2} t}{2}\right]_{0}^{\frac{\pi}{4}}-(t \tan t)_{0}^{\frac{\pi}{4}}+[\log |\sec t|]_{0}^{\frac{\pi}{4}}+\left(\frac{t^{2}}{2}\right)_{0}^{\frac{\pi}{4}} \\ & \end{aligned}

$=\left(\frac{\pi}{4}\right)^{2} \times \frac{1}{2}-\frac{\pi}{4}+\log \sqrt{2}+\left(\frac{\pi}{4}\right)^{2} \times \frac{1}{2}$

\begin{aligned} &=\frac{\pi^{2}}{16} \times \frac{1}{2}-\frac{\pi}{4}+\log \sqrt{2}+\frac{\pi^{2}}{16} \times \frac{1}{2} \\ & \end{aligned}

$=2 \times \frac{\pi^{2}}{16} \times \frac{1}{2}-\frac{\pi}{4}+\log \sqrt{2} \\$

$=\frac{\pi^{2}}{16}-\frac{\pi}{4}+\log \sqrt{2}$