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.
 $$\int \frac {1}{\sin^3(x)} dx$$
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.