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

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

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integral of x^2arccos(x)

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

\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 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

= \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})

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

Simplify

= \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

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

\int 6 sin^{6} (x) cos^{3} (x) d x

x \to 4 lim (\frac{4}{x - 4})

\int sin^{2} (x^{2}) d x

x \to - 4 - lim (\frac{2 - x}{4 - x})

x + \frac{4}{x}

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