Question
Evaluate: $\int_0^a \sqrt{a^2-x^2} d x$

Answer

Let $x=a \sin \theta$
Differentiating $ w.r.t.\ x$, we get
$d x=a \cos \theta d \theta$
Now,
$x=0$
$\Rightarrow \theta=0$
$x=a$
$\Rightarrow \theta=\frac{\pi}{2}$
$\therefore \int_0^2 \sqrt{a^2-x^2} d x$
$=\int_0^{\frac{\pi}{2}} \sqrt{a^2\left(1-\sin ^2 \theta\right)} a \cos \theta d \theta$
$=a^2 \int_0^{\frac{\pi}{2}} \cos ^2 \theta d \theta$
$=\frac{a^2}{2} \int_0^{\frac{\pi}{2}}(1+\cos 2 \theta) d \theta($using $\cos ^2 \theta=\frac{(1+\cos 2 \theta)}{2} \text { ) }$
$=\frac{a^2}{2}\left(\theta+\frac{\sin 2 \theta}{2}\right)_0^{\frac{\pi}{2}}$
$=\frac{a^2}{2}\left(\frac{\pi}{2}+0-0-0\right)$
$=\frac{\pi a^2}{4}$
$\therefore \int_0^2 \sqrt{a^2-x^2} d x=\frac{\pi a^2}{4}$

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