What is the integral of (cos^2(2θ))/(sin^4(2θ)) ?

The integral of (cos^2(2θ))/(sin^4(2θ)) is -1/6 csc^3(2θ)cos(2θ)+1/6 cot(2θ)+C

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integral of (cos^2(2θ))/(sin^4(2θ))

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

- \frac{1}{6} csc^{3} (2 θ) cos (2 θ) + \frac{1}{6} cot (2 θ) + C

Solution steps

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

Rewrite using trig identities

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

Expand

\frac{1 - sin ^{2} ( 2 θ )}{sin ^{4} ( 2 θ )} : \frac{1}{sin ^{4} ( 2 θ )} - \frac{1}{sin ^{2} ( 2 θ )}

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

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

\int \frac{1}{sin ^{4} ( 2 θ )} d θ = - \frac{1}{6} csc^{3} (2 θ) cos (2 θ) - \frac{1}{3} cot (2 θ)

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

= - \frac{1}{6} csc^{3} (2 θ) cos (2 θ) - \frac{1}{3} cot (2 θ) - (- \frac{1}{2} cot (2 θ))

Simplify

- \frac{1}{6} csc^{3} (2 θ) cos (2 θ) - \frac{1}{3} cot (2 θ) - (- \frac{1}{2} cot (2 θ)) : - \frac{1}{6} csc^{3} (2 θ) cos (2 θ) + \frac{1}{6} cot (2 θ)

= - \frac{1}{6} csc^{3} (2 θ) cos (2 θ) + \frac{1}{6} cot (2 θ)

Add a constant to the solution

= - \frac{1}{6} csc^{3} (2 θ) cos (2 θ) + \frac{1}{6} cot (2 θ) + C

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What is the integral of (cos^2(2θ))/(sin^4(2θ)) ?
The integral of (cos^2(2θ))/(sin^4(2θ)) is -1/6 csc^3(2θ)cos(2θ)+1/6 cot(2θ)+C