What is the integral of (cot^5(x))/(sin^3(x)) ?

The integral of (cot^5(x))/(sin^3(x)) is -1/7 csc^7(x)+2/5 csc^5(x)-1/3 csc^3(x)+C

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integral of (cot^5(x))/(sin^3(x))

\int \frac{cot ^{5} ( x )}{sin ^{3} ( x )} d x

- \frac{1}{7} csc^{7} (x) + \frac{2}{5} csc^{5} (x) - \frac{1}{3} csc^{3} (x) + C

Solution steps

\int \frac{cot ^{5} ( x )}{sin ^{3} ( x )} d x

Use the basic trigonometric identity:

\frac{1}{sin ( x )} = csc (x)

= \int cot^{5} (x) csc^{3} (x) d x

Apply u-substitution

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

Take the constant out:

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

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

Expand

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

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

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

\int 2 u^{4} d u = \frac{2 u ^{5}}{5}

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

= - (\frac{u ^{7}}{7} - \frac{2 u ^{5}}{5} + \frac{u ^{3}}{3})

Substitute back

u = csc (x)

= - (\frac{csc ^{7} ( x )}{7} - \frac{2 csc ^{5} ( x )}{5} + \frac{csc ^{3} ( x )}{3})

Simplify

- (\frac{csc ^{7} ( x )}{7} - \frac{2 csc ^{5} ( x )}{5} + \frac{csc ^{3} ( x )}{3}) : - \frac{1}{7} csc^{7} (x) + \frac{2}{5} csc^{5} (x) - \frac{1}{3} csc^{3} (x)

= - \frac{1}{7} csc^{7} (x) + \frac{2}{5} csc^{5} (x) - \frac{1}{3} csc^{3} (x)

Add a constant to the solution

= - \frac{1}{7} csc^{7} (x) + \frac{2}{5} csc^{5} (x) - \frac{1}{3} csc^{3} (x) + C

What is the integral of (cot^5(x))/(sin^3(x)) ?
The integral of (cot^5(x))/(sin^3(x)) is -1/7 csc^7(x)+2/5 csc^5(x)-1/3 csc^3(x)+C