Find $ \int \frac{1}{2\sin(x)-3\cos(x)}dx$. 
Find $\displaystyle \int \dfrac{1}{2\sin(x)-3\cos(x)}dx$.

My book said to solve this by saying $u = \tan \left(\dfrac{x}{2} \right)$ since $\cos(x) = \dfrac{1-u^2}{1+u^2}$ and $\sin(x) = \dfrac{2u}{1+u^2}$. I don't see how this will help since $du = \dfrac{1}{\cos(x)+1}dx $. How will we get that in the integrand?

Hint:
the suggested substitution works well because:
$$
\cos x+1=\frac{1-u^2}{1+u^2}+1=\frac{2}{1+u^2}
$$
so:
$$
du=\frac{1}{2\cos^2(x/2)}dx= \frac{dx}{\cos x+1} \Rightarrow dx=\frac{2\,du}{1+u^2}
$$
the integral becomes:
$$
2\int\frac{du}{3u^2+4u-3} 
$$
that, completing the square, becomes:
$$
2\int\frac{du}{(\sqrt{3}u+2\sqrt{3}/3)^2-13/3} 
$$
that can be solved with the substitution $t=\sqrt{3}u+2\sqrt{3}/3$ and factorizing $13/3$.
