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This question already has an answer here:

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.

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marked as duplicate by projectilemotion, C. Falcon, levap, user91500, GoodDeeds Feb 6 '17 at 8:27

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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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:

$$\int\frac{1-7\cos^2\left(x\right)}{\sin^7\left(x\right)\cos^2\left(x\right)}\space\text{d}x=\csc^6\left(x\right)\sec\left(x\right)+\text{C}\tag{5}$$

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