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  • Explain solution RD Sharma Class12 Chapter Definite Integrals exercise 19.2 question 30 maths.

Explain solution RD Sharma Class12 Chapter Definite Integrals exercise 19.2 question 30 maths.

Answers (1)

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

Hint: We use indefinite integral formula then put limits to solve this integral.

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

Solution: I=\int_{0}^{1}\frac{\tan^{-1}x}{1+x^2}dx

Put \tan^{-1}x=t

\Rightarrow \frac{1}{1+x^{2}}dx=dt

\Rightarrow dx=\left (1+x \right )^2dt

When x=0  then t=0  and when x=1  and t=\frac{\pi}{4}

 

\begin{aligned} I &=\int_{0}^{\frac{\pi}{4}} \frac{t}{1+x^{2}}\left(1+x^{2}\right) d t \\ =& \int_{0}^{\frac{\pi}{4}} t d t \\ =&\left[\frac{t^{1+1}}{1+1}\right]_{0}^{\frac{\pi}{4}} \\ =&\left[\frac{t^{2}}{2}\right]_{0}^{\frac{\pi}{4}}\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right] \end{aligned}

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

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