What is the integral of cot^6(x) ?

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

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integral of cot^6(x)

\int cot^{6} (x) d x

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

Solution steps

\int cot^{6} (x) d x

Apply u-substitution

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

Take the constant out:

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

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

Long division

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

= - \int u^{4} - u^{2} + 1 - \frac{1}{u ^{2} + 1} d u

Apply the Sum Rule:

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

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

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

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

\int 1 d u = u

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

= - (\frac{u ^{5}}{5} - \frac{u ^{3}}{3} + u - arctan (u))

Substitute back

u = cot (x)

= - (\frac{cot ^{5} ( x )}{5} - \frac{cot ^{3} ( x )}{3} + cot (x) - arctan (cot (x)))

Simplify

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

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

Add a constant to the solution

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

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