What is the integral of cot^4(5x) ?

The integral of cot^4(5x) is -1/15 cot^3(5x)+x+1/5 cot(5x)+C

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integral of cot^4(5x)

\int cot^{4} (5 x) d x

- \frac{1}{15} cot^{3} (5 x) + x + \frac{1}{5} cot (5 x) + C

Solution steps

\int cot^{4} (5 x) d x

Apply u-substitution

: \int cot^{4} (u) \frac{1}{5} d u

= \int cot^{4} (u) \frac{1}{5} d u

Take the constant out:

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

= \frac{1}{5} \cdot \int cot^{4} (u) d u

Apply Integral Reduction

= \frac{1}{5} (- \frac{cot ^{3} ( u )}{3} - \int cot^{2} (u) d u)

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

= \frac{1}{5} (- \frac{cot ^{3} ( u )}{3} - (- u - cot (u)))

Substitute back

u = 5 x

= \frac{1}{5} (- \frac{cot ^{3} ( 5 x )}{3} - (- 5 x - cot (5 x)))

Simplify

\frac{1}{5} (- \frac{cot ^{3} ( 5 x )}{3} - (- 5 x - cot (5 x))) : - \frac{1}{15} cot^{3} (5 x) + x + \frac{1}{5} cot (5 x)

= - \frac{1}{15} cot^{3} (5 x) + x + \frac{1}{5} cot (5 x)

Add a constant to the solution

= - \frac{1}{15} cot^{3} (5 x) + x + \frac{1}{5} cot (5 x) + C

\int cos (y^{4}) d y

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f (x) = (\frac{x + 8}{x - 8})^{5}

\int \frac{- e ^{\frac{1}{x}}}{x ^{2}} d x

\int (x - y) d x

What is the integral of cot^4(5x) ?
The integral of cot^4(5x) is -1/15 cot^3(5x)+x+1/5 cot(5x)+C