Evaluating $\int \frac{1-7\cos^2x}{\sin^7x\cos^2x}dx$ How do i evaluate $$\int \frac{1-7\cos^2x}{\sin^7x\cos^2x}dx$$. I tried using integration by parts and here is my approach 
$\int \frac{ sinx}{(1-cos^2x)^4\cos^2x} dx$ and then put $cos x=t$ and then tried to use partial fractions.I applied similar logic for the other part.But that made it lengthy to solve as decomposition into partial fractions is very time consuming.This question came in an objective examination in which time was limited.Can anyone help me with a shorter way to solve this problem.Thanks.
 A: Well, we know that:
$$\frac{1-7\cos^2\left(x\right)}{\sin^7\left(x\right)\cos^2\left(x\right)}=\csc^7\left(x\right)\left(\sec^2\left(x\right)-7\right)\tag1$$
So, for the integral we get:
$$\int\frac{1-7\cos^2\left(x\right)}{\sin^7\left(x\right)\cos^2\left(x\right)}\space\text{d}x=\int\csc^7\left(x\right)\sec^2\left(x\right)\space\text{d}x-7\int\csc^7\left(x\right)\space\text{d}x\tag2$$
Now, for the right integral you can use the reduction formula.
$\color{red}{\text{But}}$ using integration by parts:
$$\int\csc^7\left(x\right)\sec^2\left(x\right)\space\text{d}x=\csc^6\left(x\right)\sec\left(x\right)+7\int\csc^7\left(x\right)\space\text{d}x\tag3$$
So, we get that:
$$\int\frac{1-7\cos^2\left(x\right)}{\sin^7\left(x\right)\cos^2\left(x\right)}\space\text{d}x=\csc^6\left(x\right)\sec\left(x\right)+\color{red}{7\int\csc^7\left(x\right)\space\text{d}x-7\int\csc^7\left(x\right)\space\text{d}x}\tag4$$
Which gives that:
$$\int\frac{1-7\cos^2\left(x\right)}{\sin^7\left(x\right)\cos^2\left(x\right)}\space\text{d}x=\csc^6\left(x\right)\sec\left(x\right)+\text{C}\tag{5}$$
