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:

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

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

$= pi/4 - 1/2$

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