What is the integral of (csc^2(2x))/(cot^5(2x)) ?

The integral of (csc^2(2x))/(cot^5(2x)) is -1/2 sec^2(2x)+1/8 sec^4(2x)+1/4 sec^2(2x)+C

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integral of (csc^2(2x))/(cot^5(2x))

\int \frac{csc ^{2} ( 2 x )}{cot ^{5} ( 2 x )} d x

- \frac{1}{2} sec^{2} (2 x) + \frac{1}{8} sec^{4} (2 x) + \frac{1}{4} sec^{2} (2 x) + C

Solution steps

\int \frac{csc ^{2} ( 2 x )}{cot ^{5} ( 2 x )} d x

Rewrite using trig identities

= \int \frac{1 + cot ^{2} ( 2 x )}{cot ^{5} ( 2 x )} d x

Expand

\frac{1 + cot ^{2} ( 2 x )}{cot ^{5} ( 2 x )} : \frac{1}{cot ^{5} ( 2 x )} + \frac{1}{cot ^{3} ( 2 x )}

= \int \frac{1}{cot ^{5} ( 2 x )} + \frac{1}{cot ^{3} ( 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

= \int \frac{1}{cot ^{5} ( 2 x )} d x + \int \frac{1}{cot ^{3} ( 2 x )} d x

\int \frac{1}{cot ^{5} ( 2 x )} d x = \frac{1}{2} (ln ∣ sec (2 x) ∣ - sec^{2} (2 x) + \frac{sec ^{4} ( 2 x )}{4})

\int \frac{1}{cot ^{3} ( 2 x )} d x = - \frac{1}{2} ln ∣ sec (2 x) ∣ + \frac{sec ^{2} ( 2 x )}{4}

= \frac{1}{2} (ln ∣ sec (2 x) ∣ - sec^{2} (2 x) + \frac{sec ^{4} ( 2 x )}{4}) - \frac{1}{2} ln ∣ sec (2 x) ∣ + \frac{sec ^{2} ( 2 x )}{4}

Simplify

\frac{1}{2} (ln ∣ sec (2 x) ∣ - sec^{2} (2 x) + \frac{sec ^{4} ( 2 x )}{4}) - \frac{1}{2} ln ∣ sec (2 x) ∣ + \frac{sec ^{2} ( 2 x )}{4} : - \frac{1}{2} sec^{2} (2 x) + \frac{1}{8} sec^{4} (2 x) + \frac{1}{4} sec^{2} (2 x)

= - \frac{1}{2} sec^{2} (2 x) + \frac{1}{8} sec^{4} (2 x) + \frac{1}{4} sec^{2} (2 x)

Add a constant to the solution

= - \frac{1}{2} sec^{2} (2 x) + \frac{1}{8} sec^{4} (2 x) + \frac{1}{4} sec^{2} (2 x) + C

What is the integral of (csc^2(2x))/(cot^5(2x)) ?
The integral of (csc^2(2x))/(cot^5(2x)) is -1/2 sec^2(2x)+1/8 sec^4(2x)+1/4 sec^2(2x)+C