Question

Solve:
$I=\int \frac{1}{3\sin x-4\cos x}dx$

Answer 1

$I=\int \frac{1}{3\sin x-4\cos x}dx$

Let $3=r\cos \alpha$ and $4=r\sin \alpha$

$r^2=3^2+4^2⟹r=5$

So, $\tan \alpha =\frac{4}{3}⟹\alpha ={\tan }^{-1}\frac{4}{3}$

$3\sin x-4\cos x=r\sin x\cos \alpha -r\cos x\sin \alpha =r\sin (x-\alpha )$

$\therefore$ $I=\int \frac{1}{r\sin (x-\alpha )}dx$

$I=\frac{1}{5}\log \left(\tan \right(\frac{x}{2}-\frac{1}{2}{\tan }^{-1}\frac{4}{3}\left)\right)+c$