What is the integral of cot^4(3x) ?

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

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integral of cot^4(3x)

\int cot^{4} (3 x) d x

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

Solution steps

\int cot^{4} (3 x) d x

Apply u-substitution

: \int cot^{4} (u) \frac{1}{3} d u

= \int cot^{4} (u) \frac{1}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int cot^{4} (u) d u

Apply Integral Reduction

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

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

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

Substitute back

u = 3 x

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

Simplify

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

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

Add a constant to the solution

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

\frac{x ^{3}}{10 ( x - 2 )}

f (t) = 1 - t

\int 10 cos^{5} (x) sin^{4} (x) d x

f (x) = 3 + 8 ln (x), at x = 1

(2 x^{3}) (4 x^{2})

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