Question

Find :-
$I=\int \frac{\sin \theta }{\left(4{\cos }^2\theta \right)(2-{\sin }^2\theta )}d\theta$

Answer 1

$I=\int \frac{\sin \theta d\theta }{\left(4{\cos }^2\theta \right)(2-{\sin }^2\theta )}$

Let $\cos \theta =t$

$-\sin \theta d\theta =dt$

$\sin \theta .d\theta =-dt$

$=-\int \frac{dt}{4t^2(2-1+t^2)}$

$=-\int \frac{dt}{4t^2(1+t^2)}$

$=-\frac{1}{4}\int \frac{1+t^2-t^2}{t^2(1+t^2)}$

$=-\frac{1}{4}\int [\frac{1+t^2}{t^2(1+t^2)}-\frac{t^2}{t^2(1+t^2)}]dt$

$=-\frac{1}{4}\int [\frac{1}{t^2}-\frac{1}{1+t^2}]dt$

$=-\frac{1}{4}[-\frac{1}{t}-{\tan }^{-1}(t\left)\right]+C$

Put the value of t

$=-\frac{1}{4}[\frac{-1}{\cos \theta }-{\tan }^{-1}(\cos \theta \left)\right]+C$