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Integrate the function$\int\frac{\cos x}{\sqrt{4-\sin^2x}}$

$\begin{array}{1 1}\sin^{-1}\bigg(\large\frac{\sin x}{2}\bigg)+c \\\cos^{-1}\bigg(\large\frac{\sin x}{2}\bigg)+c \\\sin^{-1}\bigg(\large\frac{\cos x}{2}\bigg)+c \\ \cos^{-1}\bigg(\large\frac{\cos x}{2}\bigg)+c \end{array}$

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Toolbox:
• (i) if given $I=\int f(x)dx,$ let $f(x)=t,$ then $f'(x)dx=dt$,hence $\int f(x)dx=\int t.dt$
• (ii)$\int \frac{dx}{\sqrt {x^2-a^2}}=\sin ^{-1}(\frac{x}{a})+c$
Given $I=\int\frac{\cos x}{\sqrt{4-\sin^2x}}dx$

Let $\sin x=t$

on integrating we get

$\cos x dx=dt$

Hence on substituting t and dt we get,

$I=\int \large\frac{dt}{\sqrt{2^2-t^2}}$

On integrating we get,

This is of the form $\int \large\frac{dt}{\sqrt{2^2-x^2}}$

$=\sin ^{-1}(\frac{x}{a})+c$

Therefore $I=\sin^{-1}(\frac{t}{2})$

Substituting for t we get $\sin^{-1}\bigg(\large\frac{\sin x}{2}\bigg)+c$

Therefore $I=\sin^{-1}\bigg(\large\frac{\sin x}{2}\bigg)+c$

answered Feb 16, 2013 by