How do you integrate #int xarcsinx^2# by parts from #[0,1]#?

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Answer 1 · 2017-02-06T12:18:47

#= pi/4 - 1/2#

Looking first at the indefinite integral, ie

#int x arcsin(x^2) dx#

We can use IBP:

#= int (x^2/2)^prime arcsinx^2 dx#

#= x^2/2 arcsinx^2 - int x^2/2 (arcsinx^2)^prime dx#

Recognising that: #d/dx(arcsin(u)) = 1/sqrt(1 - u^2) * (du)/dx#

#= x^2/2 arcsinx^2 - int x^2/2 * (2x)/sqrt(1-x^4) dx#

#= x^2/2 arcsinx^2 - int x^3/sqrt(1-x^4) dx#

#= x^2/2 arcsinx^2 - int 2 * (-1/4) * ( sqrt(1-x^4) )^prime dx#

#= x^2/2 arcsinx^2 + 1/2 sqrt(1-x^4) + C#

The definite integral is, therefore:

#[x^2/2 arcsinx^2 + 1/2 sqrt(1-x^4) ]_0^1#

# = (1/2 * pi/2 + 1/2 * 0 ) - (0 + 1/2 ) #

#= pi/4 - 1/2#