What is the integral of x^2arcsin(sqrt(1-x^2)) ?

The integral of x^2arcsin(sqrt(1-x^2)) is -1/3 arcsin(sqrt(1-x^2))(1-x^2)^{3/2}+1/9 (-3x+x^3)+C

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integral of x^2arcsin(sqrt(1-x^2))

\int x^{2} arcsin (1 - x^{2}) d x

- \frac{1}{3} arcsin (1 - x^{2}) (1 - x^{2})^{\frac{3}{2}} + \frac{1}{9} (- 3 x + x^{3}) + C

Solution steps

\int x^{2} arcsin (1 - x^{2}) d x

Apply u-substitution

: \int - u sin^{2} (u) cos (u) d u

= \int - u sin^{2} (u) cos (u) d u

Take the constant out:

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

= - \int u sin^{2} (u) cos (u) d u

Apply Integration By Parts

: \frac{1}{3} u sin^{3} (u) - \int \frac{1}{3} sin^{3} (u) d u

= - (\frac{1}{3} u sin^{3} (u) - \int \frac{1}{3} sin^{3} (u) d u)

\int \frac{1}{3} sin^{3} (u) d u = \frac{1}{3} (- cos (u) + \frac{cos ^{3} ( u )}{3})

= - (\frac{1}{3} u sin^{3} (u) - \frac{1}{3} (- cos (u) + \frac{cos ^{3} ( u )}{3}))

Substitute back

u = arcsin (1 - x^{2})

= - (\frac{1}{3} arcsin (1 - x^{2}) sin^{3} (arcsin (1 - x^{2})) - \frac{1}{3} (- cos (arcsin (1 - x^{2})) + \frac{cos ^{3} ( arcsin ( 1 - x ^{2} ) )}{3}))

Simplify

- (\frac{1}{3} arcsin (1 - x^{2}) sin^{3} (arcsin (1 - x^{2})) - \frac{1}{3} (- cos (arcsin (1 - x^{2})) + \frac{cos ^{3} ( arcsin ( 1 - x ^{2} ) )}{3})) : - \frac{1}{3} arcsin (1 - x^{2}) (1 - x^{2})^{\frac{3}{2}} + \frac{1}{9} (- 3 x + x^{3})

= - \frac{1}{3} arcsin (1 - x^{2}) (1 - x^{2})^{\frac{3}{2}} + \frac{1}{9} (- 3 x + x^{3})

Add a constant to the solution

= - \frac{1}{3} arcsin (1 - x^{2}) (1 - x^{2})^{\frac{3}{2}} + \frac{1}{9} (- 3 x + x^{3}) + C

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What is the integral of x^2arcsin(sqrt(1-x^2)) ?
The integral of x^2arcsin(sqrt(1-x^2)) is -1/3 arcsin(sqrt(1-x^2))(1-x^2)^{3/2}+1/9 (-3x+x^3)+C