What is the integral of 6x^2arccos(x) ?

The integral of 6x^2arccos(x) is 6(1/3 x^3arccos(x)+1/3 (1/3 (1-x^2)^{3/2}-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 6x^2arccos(x)

\int 6 x^{2} arccos (x) d x

6 (\frac{1}{3} x^{3} arccos (x) + \frac{1}{3} (\frac{1}{3} (1 - x^{2})^{\frac{3}{2}} - 1 - x^{2})) + C

Solution steps

\int 6 x^{2} arccos (x) d x

Take the constant out:

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

= 6 \cdot \int x^{2} arccos (x) d x

Apply Integration By Parts

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

= 6 (\frac{1}{3} x^{3} arccos (x) - \int - \frac{x ^{3}}{3 1 - x ^{2}} d x)

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

= 6 (\frac{1}{3} x^{3} arccos (x) - (- \frac{1}{3} (\frac{1}{3} (1 - x^{2})^{\frac{3}{2}} - 1 - x^{2})))

Simplify

= 6 (\frac{1}{3} x^{3} arccos (x) + \frac{1}{3} (\frac{1}{3} (1 - x^{2})^{\frac{3}{2}} - 1 - x^{2}))

Add a constant to the solution

= 6 (\frac{1}{3} x^{3} arccos (x) + \frac{1}{3} (\frac{1}{3} (1 - x^{2})^{\frac{3}{2}} - 1 - x^{2})) + C

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

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

\int (x^{2} - 6 x + 8)^{8} (x - 3) d x

\int \frac{x ^{2} cos ( x ) - 2 x sin ( x )}{x ^{4}} d x

\int \frac{6 x}{( 3 x ^{2} - 4 ) ^{4}} d x

What is the integral of 6x^2arccos(x) ?
The integral of 6x^2arccos(x) is 6(1/3 x^3arccos(x)+1/3 (1/3 (1-x^2)^{3/2}-sqrt(1-x^2)))+C