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

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

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integral of cot^2(4x)

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

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

Solution steps

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

Rewrite using trig identities

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

\int 1 d x = x

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

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

Add a constant to the solution

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

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