# Find the antiderivative of f(x)=cos2x/(sin^2x)*(cos^2x).

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We have the function f(x) = cos 2x /(sin x)^2*(cos x)^2

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

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

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

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

=> (cosec x)^2 - (sec x)^2

So, Int [ f(x)]

=> Int [ (cosec x)^2 - (sec x)^2 dx]

=> Int [(cosec x)^2 dx] - Int [(sec x)^2 dx]

=> - tan x - cot x + C

**Therefore the required antiderivative of f(x)=cos2x/(sin^2x)*(cos^2x) = -tan x - cot x + C**

To find the anti derivative of f(x)=cos2x/(sin^2x)*(cos^2x.

Solution:

We should get Int f(x) dx = Int cos2x/(sin^2x)*(cos^2x) dx

Int f(x) dx = Int {(cos^2x- sin^2x)/(cos^xsin^2x)} dx, cos2x = cos^2x- sin^2x is an identity.

Int f(x) dx = Int (1/sin^2x) dx - Int {1/cos^2x) dx.

f(x)dx = Int {cosec^2x}dx - Int sec^2x dx dx.

f(x) = -cot x - tanx + C, as d/dx tan = sec^2x and d/dx cot x= cosecx.

**Therefore anti derivative of f(x) = cos2x/(sin^2x)*(cos^2x) = -cot x - tanx + C.**

Of course, we can determine!

You did not specified what made you to be doubtful regarding the possibility to find the anti-derivative, but I think that the cause is that the argument of the function from numerator is different from the argument of the function from denominator.

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 integrals to be additive:

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**