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Need solution for RD Sharma Maths Class 12 Chapter Definite Integrals exercise 19.2 question 54.

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Answer: a^2\left [ \frac{\pi}{4}-\frac{1}{2} \right ]  

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

Given: \int_{0}^{a} x \sqrt{\frac{a^{2}-x^{2}}{a^{2}+x^{2}}} d x


\int_{0}^{a} x \sqrt{\frac{a^{2}-x^{2}}{a^{2}+x^{2}}} d x
Put x^2=a^2\cos 2\theta \Rightarrow 2x dx=a^2-\sin2\theta .2d\theta
\Rightarrow dx=-a^2\sin 2\theta x\; d\theta
When x=0 then \theta =\frac{\pi}{4} & when x=a then \theta =0
\begin{aligned} &\therefore \int_{0}^{a} x \sqrt{\frac{a^{2}-x^{2}}{a^{2}+x^{2}}} d x=\int_{\frac{\pi}{4}}^{0} \sqrt{\frac{a^{2}(1-\cos 2 \theta)}{a^{2}(1+\cos 2 \theta)}}\left(-a^{2} \sin 2 \theta\right) d x \\ &=-a^{2} \int_{\frac{\pi}{4}}^{0} \sqrt{\frac{a^{2} 2 \sin ^{2} \theta}{a^{2} 2 \cos ^{2} \theta}}(\sin 2 \theta) d x \\ &=-a^{2} \int_{\frac{\pi}{4}}^{0} \frac{\sin \theta}{\cos \theta}(2 \sin \theta \cos \theta) d \theta \mid \end{aligned}

\begin{aligned} &=-a^{2} \int_{\frac{\pi}{4}}^{0}\left(2 \sin ^{2} \theta\right) d \theta \\ &=-a^{2} \int_{\frac{\pi}{4}}^{0}(1-\cos 2 \theta) d \theta \\ &=-a^{2} \int_{\frac{\pi}{4}}^{0} 1 d \theta+a^{2} \int_{\frac{\pi}{4}}^{0}(\cos 2 \theta) d \theta \\ &=-a^{2}[\theta]_{\frac{\pi}{4}}^{0}+a^{2}\left[\frac{\sin 2 \theta}{2}\right]_{\frac{\pi}{4}}^{0} \\ &=-a^{2}\left[0-\frac{\pi}{4}\right]+\frac{a^{2}}{2}\left[\sin 2(0)-\sin 2 \times \frac{\pi}{4}\right] \\ &=\frac{a^{2} \pi}{4}-\frac{a^{2}}{2} \cdot 1 \\ &=a^{2}\left[\frac{\pi}{4}-\frac{1}{2}\right] \end{aligned}


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