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

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

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

\int cot^{2} (\frac{4 x}{π}) d x

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

Solution steps

\int cot^{2} (\frac{4 x}{π}) d x

Apply u-substitution

: \int cot^{2} (u) \frac{π}{4} d u

= \int cot^{2} (u) \frac{π}{4} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{π}{4} \cdot \int cot^{2} (u) d u

Rewrite using trig identities

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int 1 d u = u

\int csc^{2} (u) d u = - cot (u)

= \frac{π}{4} (- u - cot (u))

Substitute back

u = \frac{4 x}{π}

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

Simplify

\frac{π}{4} (- \frac{4 x}{π} - cot (\frac{4 x}{π})) : - x - \frac{π}{4} cot (\frac{4 x}{π})

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

Add a constant to the solution

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

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