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

The integral of x(arcsin(x))^2 is

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integral of x(arcsin(x))^2

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

\frac{1}{2} x^{2} arcsin^{2} (x) - \frac{1}{2} arcsin^{2} (x) + \frac{1}{4} arcsin (x) sin (2 arcsin (x)) + \frac{1}{4} arcsin^{2} (x) + \frac{1}{8} cos (2 arcsin (x)) + C

Solution steps

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

Simplify

= \int x arcsin^{2} (x) d x

Apply Integration By Parts

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

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

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

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

Simplify

\frac{1}{2} x^{2} arcsin^{2} (x) - (\frac{1}{2} arcsin (x) (arcsin (x) - \frac{1}{2} sin (2 arcsin (x))) - \frac{1}{2} (\frac{arcsin ^{2} ( x )}{2} + \frac{1}{4} cos (2 arcsin (x)))) : \frac{1}{2} x^{2} arcsin^{2} (x) - \frac{1}{2} arcsin^{2} (x) + \frac{1}{4} arcsin (x) sin (2 arcsin (x)) + \frac{1}{4} arcsin^{2} (x) + \frac{1}{8} cos (2 arcsin (x))

= \frac{1}{2} x^{2} arcsin^{2} (x) - \frac{1}{2} arcsin^{2} (x) + \frac{1}{4} arcsin (x) sin (2 arcsin (x)) + \frac{1}{4} arcsin^{2} (x) + \frac{1}{8} cos (2 arcsin (x))

Add a constant to the solution

= \frac{1}{2} x^{2} arcsin^{2} (x) - \frac{1}{2} arcsin^{2} (x) + \frac{1}{4} arcsin (x) sin (2 arcsin (x)) + \frac{1}{4} arcsin^{2} (x) + \frac{1}{8} cos (2 arcsin (x)) + C

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