What is the integral of cot^5(6x) ?

The integral of cot^5(6x) is -1/6 (-1/2 cot^2(6x)+1/4 cot^4(6x)+1/2 ln(csc^2(6x)))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of cot^5(6x)

\int cot^{5} (6 x) d x

- \frac{1}{6} (- \frac{1}{2} cot^{2} (6 x) + \frac{1}{4} cot^{4} (6 x) + \frac{1}{2} ln (csc^{2} (6 x))) + C

Solution steps

\int cot^{5} (6 x) d x

Apply u-substitution

: \int cot^{5} (u) \frac{1}{6} d u

= \int cot^{5} (u) \frac{1}{6} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int cot^{5} (u) d u

Apply u-substitution

: \int - \frac{v ^{5}}{1 + v ^{2}} d v

= \frac{1}{6} \cdot \int - \frac{v ^{5}}{1 + v ^{2}} d v

Take the constant out:

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

= \frac{1}{6} (- \int \frac{v ^{5}}{1 + v ^{2}} d v)

Long division

\frac{v ^{5}}{1 + v ^{2}} : v^{3} - v + \frac{v}{v ^{2} + 1}

= \frac{1}{6} (- \int v^{3} - v + \frac{v}{v ^{2} + 1} 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}{6} (- (\int v^{3} d v - \int v d v + \int \frac{v}{v ^{2} + 1} d v))

\int v^{3} d v = \frac{v ^{4}}{4}

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

\int \frac{v}{v ^{2} + 1} d v = \frac{1}{2} ln v^{2} + 1

= \frac{1}{6} (- (\frac{v ^{4}}{4} - \frac{v ^{2}}{2} + \frac{1}{2} ln v^{2} + 1))

Substitute back

= \frac{1}{6} (- (\frac{cot ^{4} ( 6 x )}{4} - \frac{cot ^{2} ( 6 x )}{2} + \frac{1}{2} ln cot^{2} (6 x) + 1))

Simplify

\frac{1}{6} (- (\frac{cot ^{4} ( 6 x )}{4} - \frac{cot ^{2} ( 6 x )}{2} + \frac{1}{2} ln cot^{2} (6 x) + 1)) : - \frac{1}{6} (- \frac{1}{2} cot^{2} (6 x) + \frac{1}{4} cot^{4} (6 x) + \frac{1}{2} ln (csc^{2} (6 x)))

= - \frac{1}{6} (- \frac{1}{2} cot^{2} (6 x) + \frac{1}{4} cot^{4} (6 x) + \frac{1}{2} ln (csc^{2} (6 x)))

Add a constant to the solution

= - \frac{1}{6} (- \frac{1}{2} cot^{2} (6 x) + \frac{1}{4} cot^{4} (6 x) + \frac{1}{2} ln (csc^{2} (6 x))) + C

\frac{\partial}{\partial x} (x e^{x y} + 2 x y + \frac{1}{x})

f (x) = 2 x^{2} - 5 x + 3, at x = 0

\int \frac{8 + x + x}{x} d x

\frac{d x}{d t} = x sin (t)

x \to 1 - lim (arcsin (x))

What is the integral of cot^5(6x) ?
The integral of cot^5(6x) is -1/6 (-1/2 cot^2(6x)+1/4 cot^4(6x)+1/2 ln(csc^2(6x)))+C