Answer
$\frac{1}{5}\ln\left|\frac{2\tan\left(\frac{x}{2}\right)-1}{\tan\left(\frac{x}{2}\right)+2}\right|+C$
Work Step by Step
Substitute $t$ = $\tan\left(\frac{x}{2}\right)$
then $\frac{x}{2}$ = $\tan^{-1}t$
$\cos\frac{x}{2}$ = $\frac{1}{\sqrt {1+t^2}}$
$\sin\frac{x}{2}$ = $\frac{t}{\sqrt {1+t^2}}$
$\cos x=\frac{1-t^2}{1+t^2}$
$\sin x=\frac{2t}{1+t^2}$
$dx=\frac{2}{1+t^2}dt$
Determine the integral using the substitution:
$\int\frac{1}{3\sin x-4\cos x}dx$ = $\int\frac{1}{3\left(\frac{2t}{1+t^2}\right)-4\left(\frac{1-t^2}{1+t^2}\right)}\frac{2}{1+t^2}dt$ = $\int\frac{2}{3(2t)-4(1-t^2)}dt$ = $\int\frac{1}{2t^2+3t-2}dt$ = $\int\frac{1}{(2t-1)(t+2)}dt$ = $\frac{1}{5}\int\left(\frac{2}{2t-1}-\frac{1}{t+2}\right)dt$ = $\frac{1}{5}\left[\ln|2t-1|-\ln|t+2|\right]+C$ = $\frac{1}{5}\ln\left|\frac{2t-1}{t+2}\right|+C$ = $\frac{1}{5}\ln\left|\frac{2\tan\left(\frac{x}{2}\right)-1}{\tan\left(\frac{x}{2}\right)+2}\right|+C$
