Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XII  >>  Math  >>  Integrals
0 votes

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} $

Can you answer this question?

1 Answer

0 votes
  • (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
Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App