What is the integral of (cos(2θ))/(sin^2(2θ)) ?

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

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integral of (cos(2θ))/(sin^2(2θ))

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

\frac{1}{4} (- cot (θ) - tan (θ)) + C

Solution steps

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

Rewrite using trig identities

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

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 θ )}{cos ^{2} ( θ ) sin ^{2} ( θ )} d θ

Rewrite using trig identities

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

Expand

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

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

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

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

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

= \frac{1}{4} (- cot (θ) - tan (θ))

Add a constant to the solution

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

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