Evaluate $\int\frac{2\cos^2x+2}{\sin^3x} \, \operatorname{d}\!x$

Question

Solving the Indefinite integral: $$\int\frac{2\cos^2x+2}{\sin^3x}\, \operatorname{d}\!x$$

What's the proper way of finding Indefinite integral in this one? can't get rid of either $\cos$ or $\sin$.

Thanks.

Answer 1

Let $u=\cos x$ so $du=-\sin x dx$ and we have $$\int\frac{2\cos^2 x+2}{\sin^3x}dx=-2\int\frac{u^2+1}{(1-u^2)^2}du=-\int\frac{du}{(1-u)^2}-\int\frac{du}{(u+1)^2}\\=-(1-u)^{-1}+(1+u)^{-1}+C=\frac{1}{1+\cos x}-\frac{1}{1-\cos x}+C=-2\frac{\cos x}{\sin^2 x}+C$$

Answer 2

$$\int \frac{4-2\sin^{2}(x)}{\sin^{3}(x)} \ dx = 4\cdot \int\csc^{3}(x)\ dx -2 \int \csc(x) \ dx $$

Now you can integrate $\csc^{3}(x)$ by writing it as $\csc^{2}(x) \cdot \csc(x) = (\cot^{2}(x)-1)\cdot \csc(x)$

Answer 3

A good point for the next. Assume you have the following integral: $$ \int R(\sin x,\cos x)dx $$

If you have $$R(-\sin x,\cos x)\equiv-R(\sin x,\cos x)$$ then it is easier for us to set $\cos x=t$. And you can see this fact is satisfied for the integrand.