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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of cot^3(3x)csc^5(3x)

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

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

Solution steps

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

Apply u-substitution

= \int cot^{3} (u) csc^{5} (u) \frac{1}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int cot^{3} (u) csc^{5} (u) d u

Apply u-substitution

= \frac{1}{3} \cdot \int - v^{4} (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}{3} (- \int v^{4} (v^{2} - 1) d v)

Expand

v^{4} (v^{2} - 1) : v^{6} - v^{4}

= \frac{1}{3} (- \int v^{6} - v^{4} 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}{3} (- (\int v^{6} d v - \int v^{4} d v))

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

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

= \frac{1}{3} (- (\frac{v ^{7}}{7} - \frac{v ^{5}}{5}))

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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