What is the integral of cot^6(x)csc^4(x) ?

The integral of cot^6(x)csc^4(x) is -1/7 cot^7(x)-1/9 cot^9(x)+C

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integral of cot^6(x)csc^4(x)

\int cot^{6} (x) csc^{4} (x) d x

- \frac{1}{7} cot^{7} (x) - \frac{1}{9} cot^{9} (x) + C

Solution steps

\int cot^{6} (x) csc^{4} (x) d x

Apply u-substitution

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

Expand

u^{6} (1 + u^{2}) : u^{6} + u^{8}

= - \int u^{6} + u^{8} 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 u^{8} d u)

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

\int u^{8} d u = \frac{u ^{9}}{9}

= - (\frac{u ^{7}}{7} + \frac{u ^{9}}{9})

Substitute back

u = cot (x)

= - (\frac{cot ^{7} ( x )}{7} + \frac{cot ^{9} ( x )}{9})

Simplify

- (\frac{cot ^{7} ( x )}{7} + \frac{cot ^{9} ( x )}{9}) : - \frac{1}{7} cot^{7} (x) - \frac{1}{9} cot^{9} (x)

= - \frac{1}{7} cot^{7} (x) - \frac{1}{9} cot^{9} (x)

Add a constant to the solution

= - \frac{1}{7} cot^{7} (x) - \frac{1}{9} cot^{9} (x) + C

What is the integral of cot^6(x)csc^4(x) ?
The integral of cot^6(x)csc^4(x) is -1/7 cot^7(x)-1/9 cot^9(x)+C