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.
 $$\int \frac {1}{\sin^3(x)} dx$$
 A: 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 
A: 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. 
A: 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
$$ -\int \frac{{u^2}}{\sqrt{u^2-1}} du . $$
A: 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.
