How to integrate $\int \frac{1}{\sin^6(x)}dx$ How do I find the  $\int \dfrac{1}{\sin^6(x)}dx$?
I was able to manage till here.
I don't know how to proceed further? I want to find what will be the degree of resultant polynomial and will it be in cotx or tanx?
 A: HINT:
$$\int\dfrac{dx}{\sin^{2n+2}x}=\int\csc^{2n+2}dx=\int(1+\cot^2x)^n\csc^2x\ dx$$
A: Hint:
From the proposed solution, you were supposed to find, $$\int\dfrac1{\sin^6x}\,\mathrm{d}x=\int\dfrac{1}{\sin^4x}\,\mathrm{d}x+\int\dfrac{\cos^2x}{\sin^6x}\,\mathrm{d}x.$$ They suggested to use integration by parts for the third integral. As for the other one, you may do as follows, $$\int\dfrac{1}{\sin^4x}\,\mathrm{d}x=\int\dfrac{\sin^2x+\cos^2x}{\sin^4x}\,\mathrm{d}x=\int\dfrac{\sin^2x}{\sin^4x}\,\mathrm{d}x+\int\dfrac{\cos^2x}{\sin^4x}\,\mathrm{d}x.$$ Can you proceed further?
A: You can also use $1+cot^2(x)=cosec^2(x)$ so integral becomes $$\int (1+cot^6(x)+3cot^2(x)+3cot(x)$$ where the problem lies only in $cot^6(x)$ you will get it by reduction formula. 
A: 
You can use the reduction formula:
$$\int\csc^m(x)\space\text{d}x=-\frac{\cos(x)\csc^{m-1}(x)}{m-1}+\frac{m-2}{m-1}\int\csc^{m-2}(x)\space\text{d}x$$

So:
$$\int\frac{1}{\sin^6(x)}\space\text{d}x=\int\csc^6(x)\space\text{d}x=$$
$$-\frac{\cos(x)\csc^{6-1}(x)}{6-1}+\frac{6-2}{6-1}\int\csc^{6-2}(x)\space\text{d}x=$$
$$-\frac{\cos(x)\csc^5(x)}{5}+\frac{4}{5}\int\csc^{4}(x)\space\text{d}x$$
Now use another time the reduction formula and then notice that $\int\csc^2(x)\space\text{d}x=-\cot(x)+\text{C}$.
