What is the integral of int(sec^2x)/sqrt(4-sec^2x)dx∫sec2x√4−sec2xdx ?

int(sec^2x)/sqrt(4-sec^2x)dx∫sec2x√4−sec2xdx

1 Answer
Dreamer.N
May 12, 2018

The answer of this question=sin^(-1)(tanx/sqrt3)sin−1(tanx√3)

Explanation:

For this take tanx=t
Then sec^2x dx=dtsec2xdx=dt

Also sec^2x= 1+tan^2xsec2x=1+tan2x
Putting these value in original equation we get
intdt/(sqrt(3-t^2))∫dt√3−t2

=sin^(-1)(t/sqrt3)=sin−1(t√3)

=sin^(-1)(tanx/sqrt3)sin−1(tanx√3)

Hope it helps!!

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