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  • Explain solution RD Sharma Class12 Chapter Definite Integrals exercise 19.2 question 58 maths.

Explain solution RD Sharma Class12 Chapter Definite Integrals exercise 19.2 question 58 maths.

Answers (1)

Answer :  \frac{1}{\sqrt{2}}\left ( \tan^{-1}\sqrt{\frac{2}{3}} \right )

Hint: Use indefinite integral formula and the limits to solve this integral

Given: \int_{0}^{\frac{1}{2}} \frac{1}{\left(1+x^{2}\right) \sqrt{1-x^{2}}} d x

Solution: \int_{0}^{\frac{1}{2}} \frac{1}{\left(1+x^{2}\right) \sqrt{1-x^{2}}} d x
 put x=\sin u \Rightarrow dx=\cos \: u\; du
where x=0 then u=0 and when x=\frac{1}{2}  then u=\frac{\pi}{6}
 

Therefore,

\begin{aligned} &\int_{0}^{\frac{1}{2}} \frac{1}{\left(1+x^{2}\right) \sqrt{1-x^{2}}} d x=\int_{0}^{\frac{\pi}{6}} \frac{1}{\left(1+\sin ^{2} u\right) \sqrt{1-\sin ^{2} u}} \cos u d u \\ &=\int_{0}^{\frac{\pi}{6}} \frac{1}{\left(1+\sin ^{2} u\right) \sqrt{\cos ^{2} u}} \cos u d u \\ &=\int_{0}^{\frac{\pi}{6}} \frac{1}{\left(1+\sin ^{2} u\right) \cos u} \cos u d u \end{aligned}

\begin{aligned} &=\int_{0}^{\frac{\pi}{6}} \frac{1}{1+\sin ^{2} u} d u\\ &=\int_{0}^{\frac{\pi}{6}} \frac{\frac{1}{\cos ^{2} u}}{\frac{1+\sin ^{2} u}{\cos ^{2} u}} d u\\ &=\int_{0}^{\frac{\pi}{6}} \frac{\sec ^{2} u}{\frac{1}{\cos ^{2} u}+\frac{\sin ^{2} u}{\cos ^{2} u}} d u\\ &=\int_{0}^{\frac{\pi}{6}} \frac{\sec ^{2} u}{\sec ^{2} u+\tan ^{2} u} d u\\ &=\int_{0}^{\frac{\pi}{6}} \frac{\sec ^{2} u}{1+2 \tan ^{2} u} d u \end{aligned}

again , put \tan u=t \sec^2u\; du=dt
when u=0 then t=0 and when u=\frac{\pi}{6} then t=\frac{1}{\sqrt{3}}

Therefore ,

\begin{aligned} &\int_{0}^{\frac{1}{2}} \frac{1}{\left(1+x^{2}\right) \sqrt{1-x^{2}}} d x=\int_{0}^{\frac{\pi}{6}} \frac{\sec ^{2} u}{1+2 \tan ^{2} u} d u \\ &=\int_{0}^{\frac{1}{\sqrt{3}}} \frac{1}{1+2 t^{2}} d t \\ &=\int_{0}^{\frac{1}{\sqrt{3}}} \frac{1}{2\left(\frac{1}{2}+t^{2}\right)} d t \\ &=\frac{1}{2} \int_{0}^{\frac{1}{\sqrt{3}}} \frac{1}{\left(\frac{1}{\sqrt{2}}\right)^{2}+t^{2}} d t \end{aligned}

\begin{aligned} &=\frac{1}{2}\left[\frac{1}{\frac{1}{\sqrt{2}}} \tan ^{-1}\left(\frac{t}{\frac{1}{\sqrt{2}}}\right)\right]_{0}^{\frac{1}{\sqrt{3}}} \\ &=\frac{1}{2}[\sqrt{2}]\left[\tan ^{-1}(\sqrt{2} t)\right]_{0}^{\frac{1}{\sqrt{3}}} \\ &=\frac{1}{\sqrt{2}}\left[\tan ^{-1} \sqrt{2} \times \frac{1}{\sqrt{3}}-\tan ^{-1} \sqrt{2} \times 0\right] \\ &=\frac{1}{\sqrt{2}}\left[\tan ^{-1} \sqrt{\frac{2}{3}}-\tan ^{-1} 0\right] \\ &=\frac{1}{\sqrt{2}}\left[\tan ^{-1} \sqrt{\frac{2}{3}}-0\right] \\ &=\frac{1}{\sqrt{2}} \tan ^{-1} \sqrt{\frac{2}{3}} \end{aligned}

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