Question

What is the integral of `int(sec^2x)/sqrt(4-sec^2x)dx` ?

`int(sec^2x)/sqrt(4-sec^2x)dx`

Answer 1

The answer of this question=`sin^(-1)(tanx/sqrt3)`

Explanation:

For this take tanx=t
Then `sec^2x dx=dt`

Also `sec^2x= 1+tan^2x`
Putting these value in original equation we get
`intdt/(sqrt(3-t^2))`

`=sin^(-1)(t/sqrt3)`

=`sin^(-1)(tanx/sqrt3)`

Hope it helps!!