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

The integral of cot^4(x/6) is -2cot^3(x/6)+x+6cot(x/6)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of cot^4(x/6)

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

- 2 cot^{3} (\frac{x}{6}) + x + 6 cot (\frac{x}{6}) + C

Solution steps

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

Apply u-substitution

: \int cot^{4} (u) \cdot 6 d u

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

Take the constant out:

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

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

Apply Integral Reduction

= 6 (- \frac{cot ^{3} ( u )}{3} - \int cot^{2} (u) d u)

\int cot^{2} (u) d u = - u - cot (u)

= 6 (- \frac{cot ^{3} ( u )}{3} - (- u - cot (u)))

Substitute back

u = \frac{x}{6}

= 6 (- \frac{cot ^{3} ( \frac{x}{6} )}{3} - (- \frac{x}{6} - cot (\frac{x}{6})))

Simplify

6 (- \frac{cot ^{3} ( \frac{x}{6} )}{3} - (- \frac{x}{6} - cot (\frac{x}{6}))) : - 2 cot^{3} (\frac{x}{6}) + x + 6 cot (\frac{x}{6})

= - 2 cot^{3} (\frac{x}{6}) + x + 6 cot (\frac{x}{6})

Add a constant to the solution

= - 2 cot^{3} (\frac{x}{6}) + x + 6 cot (\frac{x}{6}) + C

\int \frac{1}{( 1 - sin ^{2} ( x ) ) ^{\frac{3}{2}}} d x

\int (\frac{sin ( x )}{2 - cos ( x )}) d x

\int (5 - 2 x)^{11} d x

\int (e^{4 t} - t^{- 1}) d t

\int (x^{3} + 2 x + 3) d x

What is the integral of cot^4(x/6) ?
The integral of cot^4(x/6) is -2cot^3(x/6)+x+6cot(x/6)+C