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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} $

1 Answer

  • (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 meena.p