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# Integral of x^2 arcsin(x) dx?

### 1 Answer

- cmcsafeLv 72 months agoFavourite answer
∫x²arcsin(x) dx =

by parts

finite f(x) = arcsinx ⇒ f'(x) = 1/√(1-x²)

differ. g'(x) = x² ⇒ g(x) = x³/3

= x³*arcsin(x)/3 - (1/3)∫x³/√(1-x²) dx =

separately evaluate the last integral

∫x³/√(1-x²) dx =

by substitution

y=√(1-x²) ⇒ y² = 1-x² ⇒ x² = 1-y² ⇒ x = √(1-y²) ⇒ x³=(1-y²)*√(1-y²)

dy = x/√(1-x²) dx = √(1-y²)/y dx ⇒ y/√(1-y²) dy = dx

= ∫[(1-y²)*√(1-y²)/y]*y/√(1-y²) dy =

= ∫(1-y²) dy =

= y³/3 - y +c =

= (1-x²)*√(1-x²)/3 - √(1-x²) + c =

= √(1-x²)*(1/3-x²/3-1) +c =

= √(1-x²)*(-x²/3-2/3) + c =

= -(1/3)*(x²+2)√(1-x²) +c

back to the integral

= x³*arcsin(x)/3 + (1/9)*(x²+2)√(1-x²) +c