What is the integral of cot^5(x)csc^6(x) ?

The integral of cot^5(x)csc^6(x) is -1/10 csc^{10}(x)+1/4 csc^8(x)-1/6 csc^6(x)+C

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integral of cot^5(x)csc^6(x)

\int cot^{5} (x) csc^{6} (x) d x

- \frac{1}{10} csc^{10} (x) + \frac{1}{4} csc^{8} (x) - \frac{1}{6} csc^{6} (x) + C

Solution steps

\int cot^{5} (x) csc^{6} (x) d x

Rewrite using trig identities

= \int \frac{( 1 - sin ^{2} ( x ) ) ^{2} cos ( x )}{sin ^{11} ( x )} d x

Apply u-substitution

: \int \frac{( 1 - u ^{2} ) ^{2}}{u ^{11}} d u

= \int \frac{( 1 - u ^{2} ) ^{2}}{u ^{11}} d u

Expand

\frac{( 1 - u ^{2} ) ^{2}}{u ^{11}} : \frac{1}{u ^{11}} - \frac{2}{u ^{9}} + \frac{1}{u ^{7}}

= \int \frac{1}{u ^{11}} - \frac{2}{u ^{9}} + \frac{1}{u ^{7}} 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 \frac{1}{u ^{11}} d u - \int \frac{2}{u ^{9}} d u + \int \frac{1}{u ^{7}} d u

\int \frac{1}{u ^{11}} d u = - \frac{1}{10 u ^{10}}

\int \frac{2}{u ^{9}} d u = - \frac{1}{4 u ^{8}}

\int \frac{1}{u ^{7}} d u = - \frac{1}{6 u ^{6}}

= - \frac{1}{10 u ^{10}} - (- \frac{1}{4 u ^{8}}) - \frac{1}{6 u ^{6}}

Substitute back

u = sin (x)

= - \frac{1}{10 sin ^{10} ( x )} - (- \frac{1}{4 sin ^{8} ( x )}) - \frac{1}{6 sin ^{6} ( x )}

Simplify

- \frac{1}{10 sin ^{10} ( x )} - (- \frac{1}{4 sin ^{8} ( x )}) - \frac{1}{6 sin ^{6} ( x )} : - \frac{1}{10} csc^{10} (x) + \frac{1}{4} csc^{8} (x) - \frac{1}{6} csc^{6} (x)

= - \frac{1}{10} csc^{10} (x) + \frac{1}{4} csc^{8} (x) - \frac{1}{6} csc^{6} (x)

Add a constant to the solution

= - \frac{1}{10} csc^{10} (x) + \frac{1}{4} csc^{8} (x) - \frac{1}{6} csc^{6} (x) + C

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What is the integral of cot^5(x)csc^6(x) ?
The integral of cot^5(x)csc^6(x) is -1/10 csc^{10}(x)+1/4 csc^8(x)-1/6 csc^6(x)+C