What is the integral of (cot^5(x))/(sin^4(x)) ?

The integral of (cot^5(x))/(sin^4(x)) is -1/6 cot^6(x)-1/8 cot^8(x)+C

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integral of (cot^5(x))/(sin^4(x))

\int \frac{cot ^{5} ( x )}{sin ^{4} ( x )} d x

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

Solution steps

\int \frac{cot ^{5} ( x )}{sin ^{4} ( x )} d x

Use the basic trigonometric identity:

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

= \int cot^{5} (x) csc^{4} (x) d x

Apply u-substitution

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

= \int - 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

= - \int u^{5} (1 + u^{2}) d u

Expand

u^{5} (1 + u^{2}) : u^{5} + u^{7}

= - \int u^{5} + 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 u^{5} d u + \int u^{7} d u)

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

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

= - (\frac{u ^{6}}{6} + \frac{u ^{8}}{8})

Substitute back

u = cot (x)

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

Simplify

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

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

Add a constant to the solution

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

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