What is the integral of cot^3(2x)csc^3(2x) ?

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

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integral of cot^3(2x)csc^3(2x)

\int cot^{3} (2 x) csc^{3} (2 x) d x

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

Solution steps

\int cot^{3} (2 x) csc^{3} (2 x) d x

Apply u-substitution

: \int cot^{3} (u) csc^{3} (u) \frac{1}{2} d u

= \int cot^{3} (u) csc^{3} (u) \frac{1}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int cot^{3} (u) csc^{3} (u) d u

Apply u-substitution

: \int - v^{2} (v^{2} - 1) d v

= \frac{1}{2} \cdot \int - v^{2} (v^{2} - 1) d v

Take the constant out:

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

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

Expand

v^{2} (v^{2} - 1) : v^{4} - v^{2}

= \frac{1}{2} (- \int v^{4} - v^{2} d v)

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (- (\int v^{4} d v - \int v^{2} d v))

\int v^{4} d v = \frac{v ^{5}}{5}

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

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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