Evaluate: x 36- ^2 x d x - Vidyadip

Question

Evaluate: $\int \frac{\sin x}{\sqrt{36-\cos ^2 x}} d x$

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Answer

$\int \frac{\sin x}{\sqrt{36-\cos ^2 x}} d x$
Substitute, cosx = t
∴ - sin x dx = dt
∴ sin x dx = - dt
The integral becomes
$\begin{aligned} & \int \frac{-d t}{\sqrt{36-t^2}} \\ & =-\int \frac{d t}{\sqrt{6^2-t^2}} \\ & =-\sin ^{-1}\left(\frac{t}{6}\right)+C \\ & =-\sin ^{-1} \cdot\left(\frac{\cos x}{6}\right)+c\end{aligned}$

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