How do I solve $\int \frac{dx}{\sin^4(x)}$?

Question

$\int \frac{dx}{\sin^4(x)}$

I tried:

$$\int \frac{dx}{\sin^4(x)} = \int \frac{dx}{(\frac{1+\cos(2x)}{2})^2} = 4\int\frac{dx}{(1+\cos(2x))^2}$$

What do I do next?

Answer 1

You could continue that route, or

$$\int \frac{1}{\sin^4 x}dx = \int \csc^4 x dx = \int \csc^2 x + \cot^2 x \csc^2 x \: dx$$

then let $u = \cot x$

$$\implies -\int 1 + u^2 \: du = -u -\frac{1}{3}u^3 + C = -\cot x - \frac{1}{3}\cot^3 x + C$$

Answer 2

Let $\csc(x)=\frac1{\sin x}.$ Then, the integral becomes $\int \csc^4(x)dx.$ Then,

$$\int \csc^4(x)dx=\int \csc^2(x)\cdot(1+\cot^2(x)) dx=\int \csc^2(x)dx+\int\cot^2(x)\csc^2(x)dx.$$

Now, letting $\cot(x)=\frac{\cos(x)}{\sin(x)},$ notice that $\frac d{dx}\cot(x)=-\csc^2(x).$ Thus, the integral above becomes:

\begin{equation} \begin{split} \int \csc^2(x)dx+\int\cot^2(x)\csc^2(x)dx&=-\int (\frac d{dx}\cot(x))dx-\int\cot^2(x)(\frac d{dx} \cot(x))dx\\ &=-\cot(x)-\frac{\cot^3(x)}3+C \end{split} \end{equation} (I used chain rule for the last equals; $\frac{d}{dx}(\cot^3(x))=3\cot^2(x)\frac d{dx}\cot(x)$)