# Evaluate the integral of function y=cos2x/cos^2x*sin^2x.

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### 2 Answers

We have y = cos 2x / (cos x)^2*(sin x)^2

y = cos 2x / (cos x)^2*(sin x)^2

=> [(cos x)^2 - (sin x)^2] / (cos x)^2*(sin x)^2

=> 1/ (sin x)^2 - 1/ (cos x)^2

Now the integral of 1/ (sin x)^2 = -1 / tan x and the intgral of 1/ (cos x)^2 is tan x

Therefore Int [1/ (sin x)^2 - 1/ (cos x)^2]

=> -cot x - tanx + C

**So the required integral is -cot x - tanx + C **

We'll re-write the numerator of the double angle:

cos 2x= (cos x)^2 - (sin x)^2

We'll re-write the indefinite integral:

Int cos2x dx/(cos x)^2*(sin x)^2 = Int {[(cos x)^2 - (sin x)^2]dx/(cos x)^2*(sin x)^2}

We'll use the property of additivity of integrals:

Int cos2x dx/(cos x)^2*(sin x)^2 = Int (cos x)^2dx/(cos x)^2*(sin x)^2 - Int (sin x)^2]dx/(cos x)^2*(sin x)^2

We'll simplify and we'll get:

Int cos2x dx/(cos x)^2*(sin x)^2 = Int dx/*(sin x)^2 - Int dx/(cos x)^2

**Int cos2x dx/(cos x)^2*(sin x)^2 = -cot x - tan x + C**