# Integral $\int \frac{\operatorname d\!x}{\sin^3 x}$

I need to calculate the following integral for my homework, but I dont know how. If someone show me step by step solution I would really appreciate it. $$\int \frac {1}{\sin^3(x)} dx$$

-
Note $1/\sin^3 x=\csc^3 x$. Evaluating your integral is similar to evaluating $\int \sec^3 x\,dx$. To see how to do this, see here. – David Mitra Dec 14 '12 at 16:23

There is a standard first-year calculus response. Rewrite the integrand as $\dfrac{\sin x}{\sin^4 x}=\dfrac{\sin x}{(1-\cos^2 x)^2}$. The substitution $u=\cos x$ leaves us integrating $-\dfrac{1}{(1-u^2)^2}$. Now partial fractions.

There are in many cases more efficient procedures, but one can in principle handle in this way all $$\int \sin^m x\cos^n x\,dx,$$ where $m$ and $n$ are integers and at least one of $m$ and $n$ is odd.

-

integral csc^3(x) dx

Use the reduction formula,

-cos(x) sin^2(x)^((m-1)/2) csc^(m-1)(x) 2F1(1/2, (m+1)/2, 3/2, cos^2(x)) = -(cos(x) csc^(m-1)(x))/(m-1) + (m-2)/(m-1)-cos(x) sin^2(x)^((m-3)/2) \csc^(m-3)(x) 2F1(1/2, (m-1)/2, 3/2, cos^2(x)), where m = 3:

= 1/2 integral csc(x) dx-1/2 cot(x) csc(x)

The integral of $\csc(x) \,\,\text{is} -\log(\cot(x)+\csc(x)):$ = $-1/2 (\cot(x) \csc(x))-1/2 \log(\cot(x)+\csc(x))+$constant

Factor the answer a different way:

Answer: $= 1/2 (-\cot(x) \csc(x)-\log(\cot(x)+\csc(x)))+\text{constant}$

Sorry for the mess.

-
This is unreadable. – Did Dec 14 '12 at 16:49
@jay It's easy to use latex, if you need to find quick mainstream LaTeX symbols, you can search them visually with this render, another useful tool that would help you to do that is Detextify, you draw the symbol and Detextify tries to guess it for you. – Voyska Dec 14 '12 at 17:07

Substitute $u=\tan\frac{x}{2}$. Then $\frac{x}{2}=\arctan u\Rightarrow x=2\arctan u$. This means $$\frac{dx}{du}=\frac{2}{1+u^2}$$ In addition, $$\sin x=\frac{2u}{u^2+1}$$ Then, $$\int\frac{1}{\sin^3x}dx=\frac{2}{8}\int\frac{(u^2+1)^2}{u^3}du=\frac{1}{4}\int u+\frac{2}{u}+\frac{1}{u^3}du$$ which I think you know how to solve.

What I used here is the classic but always useful Tangent half-angle formula

-

Make the change of variables $\csc(x) = u$, which implies

$$\int \csc(\theta)^3 d \theta = -\int \frac{{u^2}}{\sqrt{u^2-1}} du \,.$$

Following it with the change of variables $u = \sin(t)$ yields

$$-\int \frac{{u^2}}{\sqrt{u^2-1}} du = i \int \sin(t)^2\,dz= \dots.$$

Note: You can use integration by parts to evaluate the integral

$$-\int \frac{{u^2}}{\sqrt{u^2-1}} du .$$

-