What is the integral of cot^4(9x) ?

The integral of cot^4(9x) is -1/27 cot^3(9x)+x+1/9 cot(9x)+C

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integral of cot^4(9x)

\int cot^{4} (9 x) d x

- \frac{1}{27} cot^{3} (9 x) + x + \frac{1}{9} cot (9 x) + C

Solution steps

\int cot^{4} (9 x) d x

Apply u-substitution

: \int cot^{4} (u) \frac{1}{9} d u

= \int cot^{4} (u) \frac{1}{9} d u

Take the constant out:

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

= \frac{1}{9} \cdot \int cot^{4} (u) d u

Apply Integral Reduction

= \frac{1}{9} (- \frac{cot ^{3} ( u )}{3} - \int cot^{2} (u) d u)

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

= \frac{1}{9} (- \frac{cot ^{3} ( u )}{3} - (- u - cot (u)))

Substitute back

u = 9 x

= \frac{1}{9} (- \frac{cot ^{3} ( 9 x )}{3} - (- 9 x - cot (9 x)))

Simplify

\frac{1}{9} (- \frac{cot ^{3} ( 9 x )}{3} - (- 9 x - cot (9 x))) : - \frac{1}{27} cot^{3} (9 x) + x + \frac{1}{9} cot (9 x)

= - \frac{1}{27} cot^{3} (9 x) + x + \frac{1}{9} cot (9 x)

Add a constant to the solution

= - \frac{1}{27} cot^{3} (9 x) + x + \frac{1}{9} cot (9 x) + C

e^{-} x^{2}

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

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\int \frac{5 x - 2}{x ( - 2 )} d x

What is the integral of cot^4(9x) ?
The integral of cot^4(9x) is -1/27 cot^3(9x)+x+1/9 cot(9x)+C