What is the integral of 2xarccos(x) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 2xarccos(x)

\int 2 x arccos (x) d x

x^{2} arccos (x) + \frac{1}{2} arcsin (x) - \frac{1}{4} sin (2 arcsin (x)) + C

Solution steps

\int 2 x arccos (x) d x

Take the constant out:

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

= 2 \cdot \int x arccos (x) d x

Apply Integration By Parts

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

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

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

= 2 (\frac{1}{2} x^{2} arccos (x) - (- \frac{1}{4} (arcsin (x) - \frac{1}{2} sin (2 arcsin (x)))))

Simplify

2 (\frac{1}{2} x^{2} arccos (x) - (- \frac{1}{4} (arcsin (x) - \frac{1}{2} sin (2 arcsin (x))))) : x^{2} arccos (x) + \frac{1}{2} arcsin (x) - \frac{1}{4} sin (2 arcsin (x))

= x^{2} arccos (x) + \frac{1}{2} arcsin (x) - \frac{1}{4} sin (2 arcsin (x))

Add a constant to the solution

= x^{2} arccos (x) + \frac{1}{2} arcsin (x) - \frac{1}{4} sin (2 arcsin (x)) + C

\int x ln (3 x) - x d x

\int \frac{9 - 4 x ^{2}}{x ^{2}} d x

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

\int \frac{1}{24} d x

\int y^{2} sin (y^{3}) d y

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