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

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

Answers (1)

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<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}

Posted by

infoexpert27

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