Question

$I=\int \frac{\cos \theta }{(4+{\sin }^2\theta )(5-4{\cos }^2\theta )}dx$

Answer 1

$I=\int \frac{cos\theta }{(4+si{n}^2\theta )(5-4co{s}^2\theta )}d\theta$

$=\int \frac{cos\theta }{(4+si{n}^2\theta )(5-4(1-si{n}^2\theta \left)\right)}$

$=\int \frac{cos\theta }{(4+si{n}^2\theta )(1+si{n}^2\theta )}d\theta$

$=\int \frac{dm}{(4+m^2)(1+m^2)}$

$[\because sin\theta =mcos\theta d\theta =dm]$

$=\int \frac{4}{15(4m^2+1)}dm-\int \frac{1}{15(m^2+1)}dm$

$=\frac{2}{15}ta{n}^{-1}\left(2m\right)-\frac{1}{30}ta{n}^{-1}\left(\frac{m}{2}\right)+c$

replacing in with $sin\theta$

$I\Rightarrow \frac{2}{15}ta{n}^{-1}\left(2sin\theta \right)-\frac{1}{30}ta{n}^{-1}\left(\frac{sin\theta }{2}\right)+c$