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Provide solution for RD Sharma maths class 12 chapter 19 Definite Integrals exercise Fill in the blanks question 6

Answers (1)

Answer: \frac{\pi^{2}}{32}

Hint: You must know about the derivative of \tan ^{-1} x \text { and } \int x \; d x

Given: \int_{0}^{1} \frac{\tan ^{-1} x}{1+x^{2}} \mathrm{dx}

Solution:  

I=\int_{0}^{1} \frac{\tan ^{-1} x}{1+x^{2}} d x

Put 

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

When \mathrm{x}=0, \mathrm{t}=0

When \mathrm{x}=1, \mathrm{t}=\frac{\pi}{4}

    \begin{aligned} &=\int_{0}^{\pi / 4} t \; dt\\\\ &=\left[\frac{t^{2}}{2}\right]_{0}^{\pi / 4} \end{aligned}

    \begin{aligned} &=\frac{1}{2}\left[\left(\frac{\pi}{4}\right)^{2}-0\right] \\\\ &=\frac{1}{2}\left[\frac{\pi^{2}}{16}\right] \\\\ &=\frac{\pi^{2}}{32} \end{aligned}

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