$\int \frac{dx}{\sin^3{x}}$ possible with universal substitution?

Question

If function is of the form $F(\sin{x},\cos{x})dx$, then I can apply $t = \tan{\frac{x}{2}}$ substition.

Examples:

$\int \frac{1}{4+3\cos{x}}dx$

$\int \frac{\sin{x}-1}{\cos{x}+2}dx$

$\int \frac{\sin{x}}{\sin{x}+\cos{x}+1}dx$

And if function is of the form $F(\sin^2{x}, \cos^2{x}, \sin{x}\cos{x})dx$ then I can apply $t = \tan{x}$ substition.

Such as:

$\int \frac{1}{2\sin^2{x}+3\cos^2{x}} dx$

$\int \frac{1}{\cos^4{x}\sin^2{x}} dx$

Question is: what if function is of the mixed form?

1. $\int \frac{\cos^3{x}}{\sin^3{x}+\sin^2{x}}dx$

or simply:

2. $\int \frac{dx}{\sin^3{x}}$

I do realize some of those or all of those integrals can be solved in another way, however I'm only considering the universal substitution. What substitution should I consider when solving those 2 integrals?

Answer 1

Hint: For #1, $\sin^3 x + \sin^2 x = \sin^2 x (1 + \sin x)$; you can probably use a straight substitution for this, i.e. $u = \sin x, du = \cos x \ dx.$

Answer 2

Bioche's rules suggest to set

1. $u=\sin x,\;\mathrm du=\cos x\,\mathrm d x$.
2. $u=\cos x,\;\mathrm du=-\sin x\,\mathrm d x$.