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

The integral of cot^3(4x) is 1/4 (-(cot^2(4x))/2-ln|sin(4x)|)+C

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integral of cot^3(4x)

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

\frac{1}{4} (- \frac{cot ^{2} ( 4 x )}{2} - ln ∣ sin (4 x) ∣) + C

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply Integral Reduction

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

\int cot (u) d u = ln ∣ sin (u) ∣

= \frac{1}{4} (- \frac{cot ^{2} ( u )}{2} - ln ∣ sin (u) ∣)

Substitute back

u = 4 x

= \frac{1}{4} (- \frac{cot ^{2} ( 4 x )}{2} - ln ∣ sin (4 x) ∣)

Add a constant to the solution

= \frac{1}{4} (- \frac{cot ^{2} ( 4 x )}{2} - ln ∣ sin (4 x) ∣) + C

f (x) = \frac{4 x}{x + 2}, at x = 6

\frac{d}{d x} (3 x + 8)

\int cos^{2} (x t a) n^{3} x d x

3 u + 5 u

y^{^{'}} = 4 - 9^{2} - 6 x^{5}

What is the integral of cot^3(4x) ?
The integral of cot^3(4x) is 1/4 (-(cot^2(4x))/2-ln|sin(4x)|)+C