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

The integral of cot^5(x)csc(x) is -1/5 csc^5(x)+2/3 csc^3(x)-csc(x)+C

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

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

- \frac{1}{5} csc^{5} (x) + \frac{2}{3} csc^{3} (x) - csc (x) + C

Solution steps

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

Rewrite using trig identities

= \int \frac{( 1 - sin ^{2} ( x ) ) ^{2} cos ( x )}{sin ^{6} ( x )} d x

Apply u-substitution

: \int \frac{( 1 - u ^{2} ) ^{2}}{u ^{6}} d u

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

Expand

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

= \int \frac{1}{u ^{6}} - \frac{2}{u ^{4}} + \frac{1}{u ^{2}} 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 ^{6}} d u - \int \frac{2}{u ^{4}} d u + \int \frac{1}{u ^{2}} d u

\int \frac{1}{u ^{6}} d u = - \frac{1}{5 u ^{5}}

\int \frac{2}{u ^{4}} d u = - \frac{2}{3 u ^{3}}

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

= - \frac{1}{5 u ^{5}} - (- \frac{2}{3 u ^{3}}) - \frac{1}{u}

Substitute back

u = sin (x)

= - \frac{1}{5 sin ^{5} ( x )} - (- \frac{2}{3 sin ^{3} ( x )}) - \frac{1}{sin ( x )}

Simplify

- \frac{1}{5 sin ^{5} ( x )} - (- \frac{2}{3 sin ^{3} ( x )}) - \frac{1}{sin ( x )} : - \frac{1}{5} csc^{5} (x) + \frac{2}{3} csc^{3} (x) - csc (x)

= - \frac{1}{5} csc^{5} (x) + \frac{2}{3} csc^{3} (x) - csc (x)

Add a constant to the solution

= - \frac{1}{5} csc^{5} (x) + \frac{2}{3} csc^{3} (x) - csc (x) + C

\int \frac{x ^{2} - x + 28}{x ^{3} + 7 x} d x

\int \frac{- 2 x ^{2} + 8 x - 20}{x ^{2} - 4 x + 8} d x

n = 1 \sum \infty \frac{1}{1 + 3 ^{n}}

x d y = 4 y d x

What is the integral of cot^5(x)csc(x) ?
The integral of cot^5(x)csc(x) is -1/5 csc^5(x)+2/3 csc^3(x)-csc(x)+C