What is the integral of xarcsin(x)rcsin(x) ?

The integral of xarcsin(x)rcsin(x) is

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integral of xarcsin(x)rcsin(x)

\int x arcsin (x) d arcsin (x) d x

d (\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) d arcsin (x) d x

Take the constant out:

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

= d \cdot \int x arcsin (x) arcsin (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

= d (\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)))

= d (\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

d (\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))))) : d (\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)))

= d (\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

= d (\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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What is the integral of xarcsin(x)rcsin(x) ?
The integral of xarcsin(x)rcsin(x) is