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Evaluate : \[ \int_{-a}^a \sqrt{\frac{a-x}{a+x}}dx \]

$\begin{array}{1 1} a\pi \\ \pi \\ 3 \pi \\ 0\end{array} $

1 Answer

Solution :
Let $I= \int \limits_{-a}^a \sqrt {\Large\frac{a-x}{a+x}}$$dx= \int \limits_{-a}^a \sqrt{ \large\frac{(a-x)(a-x)}{(a+x)(a-x)}}$$dx$
$\qquad= \int\limits_{-a}^a \large\frac{a-x}{a^2-x^2}$$dx$
$\qquad= a \int \limits_{-a}^a \large\frac{dx}{\sqrt{a^2-x^2}}$$- \int \limits_{-a}^a \large\frac{x}{\sqrt{a^2-x^2}}$$dx-2a \int\limits_{-a}^a \large\frac{dx}{\sqrt{a^2-x^2}}$$-0$
$\qquad= 2a \int \limits_0^a \large\frac{dx}{\sqrt{a^2-x^2}}$$=2a \bigg[\sin^{-1} \large\frac{x}{a} \bigg]_0^a$
$\qquad= 2a\bigg[\sin ^{-1}\large\frac{x}{a}\bigg]_0^x$
$\qquad= 2a[\sin ^{-1}1 - \sin ^{-1}0]$
$\qquad= 2a \bigg(\large\frac{\pi}{2} -0 \bigg)=a \pi$
answered Feb 16 by meena.p
 
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