What is the integral of (cot^2(4x))/(csc(4x)) ?

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

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integral of (cot^2(4x))/(csc(4x))

\int \frac{cot ^{2} ( 4 x )}{csc ( 4 x )} d x

\frac{1}{4} cos (4 x) + \frac{1}{4} ln ∣ tan (2 x) ∣ + C

Solution steps

\int \frac{cot ^{2} ( 4 x )}{csc ( 4 x )} d x

Rewrite using trig identities

= \int \frac{- 1 + csc ^{2} ( 4 x )}{csc ( 4 x )} d x

Expand

\frac{- 1 + csc ^{2} ( 4 x )}{csc ( 4 x )} : - \frac{1}{csc ( 4 x )} + csc (4 x)

= \int - \frac{1}{csc ( 4 x )} + csc (4 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}{csc ( 4 x )} d x + \int csc (4 x) d x

\int \frac{1}{csc ( 4 x )} d x = - \frac{1}{4} cos (4 x)

\int csc (4 x) d x = \frac{1}{4} ln ∣ tan (2 x) ∣

= - (- \frac{1}{4} cos (4 x)) + \frac{1}{4} ln ∣ tan (2 x) ∣

Simplify

= \frac{1}{4} cos (4 x) + \frac{1}{4} ln ∣ tan (2 x) ∣

Add a constant to the solution

= \frac{1}{4} cos (4 x) + \frac{1}{4} ln ∣ tan (2 x) ∣ + C

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