What is the integral of cot^3(x)csc^6(x) ?

The integral of cot^3(x)csc^6(x) is -1/4 cot^4(x)-1/3 cot^6(x)-1/8 cot^8(x)+C

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integral of cot^3(x)csc^6(x)

\int cot^{3} (x) csc^{6} (x) d x

- \frac{1}{4} cot^{4} (x) - \frac{1}{3} cot^{6} (x) - \frac{1}{8} cot^{8} (x) + C

Solution steps

\int cot^{3} (x) csc^{6} (x) d x

Apply u-substitution

: \int - w^{3} (1 + w^{2})^{2} d w

= \int - w^{3} (1 + w^{2})^{2} d w

Take the constant out:

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

= - \int w^{3} (1 + w^{2})^{2} d w

Expand

w^{3} (1 + w^{2})^{2} : w^{3} + 2 w^{5} + w^{7}

= - \int w^{3} + 2 w^{5} + w^{7} 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

= - (\int w^{3} d w + \int 2 w^{5} d w + \int w^{7} d w)

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

\int 2 w^{5} d w = \frac{w ^{6}}{3}

\int w^{7} d w = \frac{w ^{8}}{8}

= - (\frac{w ^{4}}{4} + \frac{w ^{6}}{3} + \frac{w ^{8}}{8})

Substitute back

w = cot (x)

= - (\frac{cot ^{4} ( x )}{4} + \frac{cot ^{6} ( x )}{3} + \frac{cot ^{8} ( x )}{8})

Simplify

- (\frac{cot ^{4} ( x )}{4} + \frac{cot ^{6} ( x )}{3} + \frac{cot ^{8} ( x )}{8}) : - \frac{1}{4} cot^{4} (x) - \frac{1}{3} cot^{6} (x) - \frac{1}{8} cot^{8} (x)

= - \frac{1}{4} cot^{4} (x) - \frac{1}{3} cot^{6} (x) - \frac{1}{8} cot^{8} (x)

Add a constant to the solution

= - \frac{1}{4} cot^{4} (x) - \frac{1}{3} cot^{6} (x) - \frac{1}{8} cot^{8} (x) + C

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What is the integral of cot^3(x)csc^6(x) ?
The integral of cot^3(x)csc^6(x) is -1/4 cot^4(x)-1/3 cot^6(x)-1/8 cot^8(x)+C