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

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

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integral of cot^4(2t)

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

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

Solution steps

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

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 t

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (\frac{( x ^{2} - 4 ) ^{3}}{x ^{2} + 2})

f (X) = ln (ln (ln (tan (X))))

\frac{\partial}{\partial x} (y \cdot ln (x) + \frac{1}{9} y^{3})

\frac{\partial}{\partial x} (3 x^{2})

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

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