What is the integral of (cos(2x))/(1-cos^2(2x)) ?

The integral of (cos(2x))/(1-cos^2(2x)) is 1/4 (-cot(x)-tan(x))+C

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integral of (cos(2x))/(1-cos^2(2x))

\int \frac{cos ( 2 x )}{1 - cos ^{2} ( 2 x )} d x

\frac{1}{4} (- cot (x) - tan (x)) + C

Solution steps

\int \frac{cos ( 2 x )}{1 - cos ^{2} ( 2 x )} d x

Simplify

\frac{cos ( 2 x )}{1 - cos ^{2} ( 2 x )} : \frac{cos ( 2 x )}{sin ^{2} ( 2 x )}

= \int \frac{cos ( 2 x )}{sin ^{2} ( 2 x )} d x

Rewrite using trig identities

= \int \frac{cos ( 2 x )}{4 cos ^{2} ( x ) sin ^{2} ( x )} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{4} \cdot \int \frac{cos ( 2 x )}{cos ^{2} ( x ) sin ^{2} ( x )} d x

Rewrite using trig identities

= \frac{1}{4} \cdot \int \frac{cos ^{2} ( x ) - sin ^{2} ( x )}{cos ^{2} ( x ) sin ^{2} ( x )} d x

Expand

\frac{cos ^{2} ( x ) - sin ^{2} ( x )}{cos ^{2} ( x ) sin ^{2} ( x )} : \frac{1}{sin ^{2} ( x )} - \frac{1}{cos ^{2} ( x )}

= \frac{1}{4} \cdot \int \frac{1}{sin ^{2} ( x )} - \frac{1}{cos ^{2} ( x )} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{4} (\int \frac{1}{sin ^{2} ( x )} d x - \int \frac{1}{cos ^{2} ( x )} d x)

\int \frac{1}{sin ^{2} ( x )} d x = - cot (x)

\int \frac{1}{cos ^{2} ( x )} d x = tan (x)

= \frac{1}{4} (- cot (x) - tan (x))

Add a constant to the solution

= \frac{1}{4} (- cot (x) - tan (x)) + C

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What is the integral of (cos(2x))/(1-cos^2(2x)) ?
The integral of (cos(2x))/(1-cos^2(2x)) is 1/4 (-cot(x)-tan(x))+C