What is the integral of cot^6(2v) ?

The integral of cot^6(2v) is -1/2 (1/5 cot^5(2v)-1/3 cot^3(2v)+cot(2v)-arctan(cot(2v)))+C

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integral of cot^6(2v)

\int cot^{6} (2 v) d v

- \frac{1}{2} (\frac{1}{5} cot^{5} (2 v) - \frac{1}{3} cot^{3} (2 v) + cot (2 v) - arctan (cot (2 v))) + C

Solution steps

\int cot^{6} (2 v) d v

Apply u-substitution

: \int cot^{6} (u) \frac{1}{2} d u

= \int cot^{6} (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^{6} (u) d u

Apply u-substitution

: \int - \frac{w ^{6}}{1 + w ^{2}} d w

= \frac{1}{2} \cdot \int - \frac{w ^{6}}{1 + w ^{2}} d w

Take the constant out:

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

= \frac{1}{2} (- \int \frac{w ^{6}}{1 + w ^{2}} d w)

Long division

\frac{w ^{6}}{1 + w ^{2}} : w^{4} - w^{2} + 1 - \frac{1}{w ^{2} + 1}

= \frac{1}{2} (- \int w^{4} - w^{2} + 1 - \frac{1}{w ^{2} + 1} d w)

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (- (\int w^{4} d w - \int w^{2} d w + \int 1 d w - \int \frac{1}{w ^{2} + 1} d w))

\int w^{4} d w = \frac{w ^{5}}{5}

\int w^{2} d w = \frac{w ^{3}}{3}

\int 1 d w = w

\int \frac{1}{w ^{2} + 1} d w = arctan (w)

= \frac{1}{2} (- (\frac{w ^{5}}{5} - \frac{w ^{3}}{3} + w - arctan (w)))

Substitute back

= \frac{1}{2} (- (\frac{cot ^{5} ( 2 v )}{5} - \frac{cot ^{3} ( 2 v )}{3} + cot (2 v) - arctan (cot (2 v))))

Simplify

\frac{1}{2} (- (\frac{cot ^{5} ( 2 v )}{5} - \frac{cot ^{3} ( 2 v )}{3} + cot (2 v) - arctan (cot (2 v)))) : - \frac{1}{2} (\frac{1}{5} cot^{5} (2 v) - \frac{1}{3} cot^{3} (2 v) + cot (2 v) - arctan (cot (2 v)))

= - \frac{1}{2} (\frac{1}{5} cot^{5} (2 v) - \frac{1}{3} cot^{3} (2 v) + cot (2 v) - arctan (cot (2 v)))

Add a constant to the solution

= - \frac{1}{2} (\frac{1}{5} cot^{5} (2 v) - \frac{1}{3} cot^{3} (2 v) + cot (2 v) - arctan (cot (2 v))) + C

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What is the integral of cot^6(2v) ?
The integral of cot^6(2v) is -1/2 (1/5 cot^5(2v)-1/3 cot^3(2v)+cot(2v)-arctan(cot(2v)))+C