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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= 4 \cdot \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

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

Expand

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

= 4 (- \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

= 4 (- (\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}

= 4 (- (\frac{u ^{7}}{7} + \frac{u ^{9}}{9}))

Substitute back

u = cot (x)

= 4 (- (\frac{cot ^{7} ( x )}{7} + \frac{cot ^{9} ( x )}{9}))

Simplify

4 (- (\frac{cot ^{7} ( x )}{7} + \frac{cot ^{9} ( x )}{9})) : - 4 (\frac{1}{7} cot^{7} (x) + \frac{1}{9} cot^{9} (x))

= - 4 (\frac{1}{7} cot^{7} (x) + \frac{1}{9} cot^{9} (x))

Add a constant to the solution

= - 4 (\frac{1}{7} cot^{7} (x) + \frac{1}{9} cot^{9} (x)) + C

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