What is the integral of 9cot^5(x) ?

The integral of 9cot^5(x) is -9(-1/2 cot^2(x)+1/4 cot^4(x)+1/2 ln(csc^2(x)))+C

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integral of 9cot^5(x)

\int 9 cot^{5} (x) d x

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

Solution steps

\int 9 cot^{5} (x) d x

Take the constant out:

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

= 9 \cdot \int cot^{5} (x) d x

Apply u-substitution

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

= 9 \cdot \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

= 9 (- \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}

= 9 (- \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

= 9 (- (\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

= 9 (- (\frac{u ^{4}}{4} - \frac{u ^{2}}{2} + \frac{1}{2} ln u^{2} + 1))

Substitute back

u = cot (x)

= 9 (- (\frac{cot ^{4} ( x )}{4} - \frac{cot ^{2} ( x )}{2} + \frac{1}{2} ln cot^{2} (x) + 1))

Simplify

9 (- (\frac{cot ^{4} ( x )}{4} - \frac{cot ^{2} ( x )}{2} + \frac{1}{2} ln cot^{2} (x) + 1)) : - 9 (- \frac{1}{2} cot^{2} (x) + \frac{1}{4} cot^{4} (x) + \frac{1}{2} ln (csc^{2} (x)))

= - 9 (- \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

= - 9 (- \frac{1}{2} cot^{2} (x) + \frac{1}{4} cot^{4} (x) + \frac{1}{2} ln (csc^{2} (x))) + C

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What is the integral of 9cot^5(x) ?
The integral of 9cot^5(x) is -9(-1/2 cot^2(x)+1/4 cot^4(x)+1/2 ln(csc^2(x)))+C