Question

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

Answer 1

$\int \frac{\sin \theta d\theta }{(4+{\cos }^2\theta )(2-{\sin }^2\theta )}$

Lety $x=\cos \theta \Rightarrow dx=-\sin \theta d\theta$

$-\int \frac{dx}{(4+x^2)(1+x^2)}$ $|\therefore 2{\sin }^2\theta =1+(1-{\sin }^2\theta )=1+{\cos }^2\theta =1+x^2$

$=-\frac{1}{3}\int (\frac{1}{1+x^2}-\frac{1}{4+x^2})dx$

$=\frac{-1}{3}[{\tan }^{-1}x-\frac{1}{2}{\tan }^{-1}(\frac{x}{2}\left)\right]$

$=\frac{1}{3}\left[{\tan }^{-1}\right(\cos \theta )-\frac{1}{2}{\tan }^{-1}(\frac{\cos \theta }{2}\left)\right]$