Answer
$ \int\frac{1}{3sinx -4cosx}dx=\frac{1}{5}\ln|2(tan\frac{x}{2})-1|-\frac{1}{5}\ln|tan\frac{x}{2}+2|+C$
Work Step by Step
$Ɪ=\int\frac{1}{3sinx -4cosx}dx$
Using the substitution $t=tan\frac{x}{2}$
then: $sinx =\frac{2t}{1+t^{2}}$ and $cosx=\frac{1-t^{2}}{1+t^{2}}$
and $dx=\frac{2}{1+t^{2}}dt$
then the integral becomes
Ɪ=$\int\frac{1}{3(\frac{2t}{1+t^{2}}) -4(\frac{1-t^{2}}{1+t^{2}})}\frac{2}{1+t^{2}}dt$
$ Ɪ=\int\frac{1}{2t^{2}+3t-2}dt=\int\frac{1}{(2t-1)(t+2)}dt$
the integrand is a partial fraction can be written as:
$\frac{1}{2t^{2}+3t-2}=\frac{A}{2t-1}+\frac{B}{t+2}$
Multiplying by $(2t-1)(t+2)$
$A(t+2)+B(2t-1)=1$
Or
$2A-B=1$
and
$A+2B=0$
solving the equation we found that
$A=\frac{2}{5}$ and $B=\frac{-1}{5}$
then the integral is
$ Ɪ=\int\frac{1}{(2t-1)(t+2)}dt=\frac{2}{5}\int\frac{dt}{2t-1}-\frac{1}{5}\int\frac{dt}{t+2}$
substituting $u=2t-1$ in the first integral we find
$\int\frac{dt}{2t-1}=\frac{1}{2}\int\frac{1}{u}=\frac{1}{2}\ln|u|+C=\frac{1}{2}\ln|2t-1|+C1$
the second integral is
$\int\frac{dt}{t+2}=\ln|t+2|+C2$
then
$ Ɪ=\frac{2}{5}\int\frac{dt}{2t-1}-\frac{1}{5}\int\frac{dt}{t+2}=\frac{1}{5}\ln|2t-1|-\frac{1}{5}\ln|t+2|+C$
since $t=tan\frac{x}{2}$
then
$ Ɪ=\frac{1}{5}\ln|2(tan\frac{x}{2})-1|-\frac{1}{5}\ln|tan\frac{x}{2}+2|+C$