#### Need solution for RD Sharma Maths Class 12 Chapter Definite Integrals exercise 19.2 question 55.

Answer: $\pi a$

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

Given: $\int_{-a}^{a}\sqrt{\frac{a-x}{a+x}}dx$

Solution:

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

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

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