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

The integral of cot^4(2x) is -1/6 cot^3(2x)+x+1/2 cot(2x)+C

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integral of cot^4(2x)

\int cot^{4} (2 x) d x

- \frac{1}{6} cot^{3} (2 x) + x + \frac{1}{2} cot (2 x) + C

Solution steps

\int cot^{4} (2 x) d x

Apply u-substitution

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

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

Take the constant out:

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

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

Apply Integral Reduction

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

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

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

Substitute back

u = 2 x

= \frac{1}{2} (- \frac{cot ^{3} ( 2 x )}{3} - (- 2 x - cot (2 x)))

Simplify

\frac{1}{2} (- \frac{cot ^{3} ( 2 x )}{3} - (- 2 x - cot (2 x))) : - \frac{1}{6} cot^{3} (2 x) + x + \frac{1}{2} cot (2 x)

= - \frac{1}{6} cot^{3} (2 x) + x + \frac{1}{2} cot (2 x)

Add a constant to the solution

= - \frac{1}{6} cot^{3} (2 x) + x + \frac{1}{2} cot (2 x) + C

\frac{x}{1 - ln ( x - 4 )}

f (x) = 2 x^{2} - 3

x - sin (x)

\frac{\partial}{\partial x} (sin (\frac{2 y}{x}))

\frac{d}{d x} (\frac{6}{( 1 + x ) ^{4}})

What is the integral of cot^4(2x) ?
The integral of cot^4(2x) is -1/6 cot^3(2x)+x+1/2 cot(2x)+C