Solve indefinite integral $\int\frac{dx}{\sin^2{x}\cos^3{x}}$

Question

I need to solve the integral below
$$
\int\frac{dx}{\sin^2{x}\cos^3{x}}
$$
without using hyperbolic functions but using substitutions like $u=\tan{x}$, $u=\sin{x}$ or $u=\cos{x}$.
Also, I know the correct answer:
$$
\frac{3}{2}\ln\left | \tan{\frac{x}{2}+\frac{\pi}{4}} \right | – \frac{1}{\sin{x}}+\frac{\sin{x}}{2\cos^2{x}}
$$
For me it looks like it will make sense to set $u=\sin{x}$ as long as $R(\sin{x}; -\cos{x}) = -R(\sin{x}; \cos{x})$. However, such substitution doesn't give me the expected result.

Answer 1

Note that\begin{align}\int\frac{\mathrm dx}{\sin^2x\cos^3x}&=\int\frac{\cos x}{\sin^2x\cos^4x}\,\mathrm dx\\&=\int\frac{\cos x}{\sin^2x(1-\sin^2x)^2}\,\mathrm dx.\end{align}So, by doing $\sin x=y$ and $\cos x\,\mathrm dx=\mathrm dy$, your indefinite integral becomes$$\int\frac{\mathrm dy}{y^2(1-y^2)^2}.$$Now, use the fact that\begin{align}\frac1{y^2(1-y^2)^2}&=\frac1{(1-y^2)^2}+\frac1{1-y^2}+\frac1{y^2}\\&=\frac3{4(y+1)}+\frac1{4(y+1)^2}-\frac3{4 (y-1)}+\frac1{4 (y-1)^2}+\frac1{y^2}.\end{align}