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

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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


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


\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<t<\frac{\pi}{4}

\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}

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