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

The integral of 6x^2*arccos(x) is 6(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 6x^2*arccos(x)

\int 6 x^{2} \cdot 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

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