What is the integral of cot^5(x) ?

The integral of cot^5(x) is 1/2 cot^2(x)-1/4 cot^4(x)-1/2 ln(csc^2(x))+C

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integral of cot^5(x)

\int cot^{5} (x) d x

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

Solution steps

\int cot^{5} (x) d x

Apply u-substitution

= \int - \frac{u ^{5}}{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 ^{5}}{1 + u ^{2}} d u

Long division

\frac{u ^{5}}{1 + u ^{2}} : u^{3} - u + \frac{u}{u ^{2} + 1}

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

\int u^{3} d u = \frac{u ^{4}}{4}

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

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

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

Substitute back

u = cot (x)

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

Simplify

- (\frac{cot ^{4} ( x )}{4} - \frac{cot ^{2} ( x )}{2} + \frac{1}{2} ln cot^{2} (x) + 1) : \frac{1}{2} cot^{2} (x) - \frac{1}{4} cot^{4} (x) - \frac{1}{2} ln (csc^{2} (x))

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

Add a constant to the solution

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

What is the integral of cot^5(x) ?
The integral of cot^5(x) is 1/2 cot^2(x)-1/4 cot^4(x)-1/2 ln(csc^2(x))+C