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Please Solve RD Sharma Class 12 Chapter 19 Definite Integrals Exercise Revision Exercise Question 5 Maths Textbook Solution.

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Answer:  \frac{\pi}{4}-\frac{1}{2} \log 2

Given:  \int_{0}^{1} \tan ^{-1} x d x

Hint: Use the integration by parts method


\int_{0}^{1} \tan ^{-1} x d x

\int_{0}^{1} \tan ^{-1} x \cdot 1 d x

We know that  \int u v d x=u \int v d x-v \int u d x

\begin{aligned} &=\left(x\tan ^{-1} x \right)_{0}^{1}-\int_{0}^{1} \frac{1}{1+x^{2}} \times x d x \\ & \end{aligned}

=\frac{\pi}{4}-\frac{1}{2} \int_{0}^{1} \frac{1}{t} d t \\

=\frac{\pi}{4}-\frac{1}{2}[\log (t)]_{0}^{1}=\frac{\pi}{4}-\frac{1}{2}\left[\log \left(1+x^{2}\right)\right]_{0}^{1}

\begin{aligned} &=\frac{\pi}{4}-\frac{1}{2}(\log 2-\log 1) \\ & \end{aligned}                                       (\because log 1= 0)

=\frac{\pi}{4}-\frac{1}{2} \log 2

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