Mathematics

$$\underset{-a}{\overset{a}{\int}} \sqrt{\dfrac{a - x}{a + x}} dx$$ is equal to 


ANSWER

$$a \pi$$


SOLUTION
$$I=\int^a_{-a}\sqrt{\dfrac{a-x}{a+x}}dx$$
Put $$x=a\cos\theta $$
$$dx=-a\sin d\theta$$
$$ x\Rightarrow -a=a \cos\theta \Rightarrow \theta =\pi$$
$$x\Rightarrow a=\cos\theta\Rightarrow \theta =0$$
$$\displaystyle \int^{0}_{\pi}\sqrt{\frac{a-a \cos\theta}{a+a cos\theta}}\times (-a\sin\theta)d\theta$$

$$\displaystyle \int ^{\pi}_0\sqrt{\frac{a (1-\cos\theta}{a(1+cos\theta)}}\times a\sin\theta d\theta$$
$$\displaystyle \int^{\pi}_0\sqrt{\frac{2\sin^2\frac{\theta}{2}}{2\cos^2\frac{\theta}{2}}}\times a\sin\theta d\theta$$
$$\displaystyle a\int^{\pi}_0\tan\frac{\theta}{2}\times 2\sin\frac{\theta}{2}\cos\frac{\theta}{2}d\theta$$.

$$\displaystyle \Rightarrow 2a\int^{\pi}_0\sin^2\frac{\theta}{2}d\theta$$
$$\displaystyle \Rightarrow a\int^{\pi}_0(1-\cos\theta)d\theta$$
$$\displaystyle \Rightarrow a\,\,\,\theta]^{\pi}_0+a\sin]^{\pi}_0$$
$$\Rightarrow a( \pi-0)+a(0-0)$$
$$\Rightarrow \boxed{\pi a}$$
$$\therefore \boxed{ \int^a_{-a}\sqrt{\frac{a-x}{a+x}}dx=\pi a}$$
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Single Correct Medium Published on 17th 09, 2020
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