What is the integral of 4arccos(x) ?

The integral of 4arccos(x) is 4(xarccos(x)-sqrt(1-x^2))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 4arccos(x)

\int 4 arccos (x) d x

4 (x arccos (x) - 1 - x^{2}) + C

Solution steps

\int 4 arccos (x) d x

Take the constant out:

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

= 4 \cdot \int arccos (x) d x

Apply Integration By Parts

: x arccos (x) - \int - \frac{x}{1 - x ^{2}} d x

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

\int - \frac{x}{1 - x ^{2}} d x = 1 - x^{2}

= 4 (x arccos (x) - 1 - x^{2})

Add a constant to the solution

= 4 (x arccos (x) - 1 - x^{2}) + C

- \frac{5}{13} \frac{d θ}{d t} = \frac{1}{13} (5)

\int (\frac{2 x + 4}{x ^{2} + 4 x + 8}) d x

\int \frac{3}{2 - x - x ^{2}} d x

\int_{0}^{\frac{π}{4}} cos (6 x) d x

y^{^{'}} (e^{x} - 1) ln (y) = y

What is the integral of 4arccos(x) ?
The integral of 4arccos(x) is 4(xarccos(x)-sqrt(1-x^2))+C