What is the integral of (cot^2(x)+cot^4(x)) ?

The integral of (cot^2(x)+cot^4(x)) is -1/3 cot^3(x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (cot^2(x)+cot^4(x))

\int (cot^{2} (x) + cot^{4} (x)) d x

- \frac{1}{3} cot^{3} (x) + C

Solution steps

\int cot^{2} (x) + cot^{4} (x) d x

Apply the Sum Rule:

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

= \int cot^{2} (x) d x + \int cot^{4} (x) d x

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

\int cot^{4} (x) d x = - \frac{1}{3} cot^{3} (x) + x + cot (x)

= - x - cot (x) - \frac{1}{3} cot^{3} (x) + x + cot (x)

Simplify

= - \frac{1}{3} cot^{3} (x)

Add a constant to the solution

= - \frac{1}{3} cot^{3} (x) + C

\frac{d}{d x} (ln (\frac{x + 2}{x - 2}))

y = 6 x^{4} ln (x)

f (x) = 5 x, at x = 5

\int tan (5 x) sec^{2} (5 x) d x

x - 2, x, [0, 4]

What is the integral of (cot^2(x)+cot^4(x)) ?
The integral of (cot^2(x)+cot^4(x)) is -1/3 cot^3(x)+C