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

Feb 6, 2017

$= \frac{\pi}{4} - \frac{1}{2}$

Explanation:

Looking first at the indefinite integral, ie

$\int x \arcsin \left({x}^{2}\right) \mathrm{dx}$

We can use IBP:

$= \int {\left({x}^{2} / 2\right)}^{p} r i m e \arcsin {x}^{2} \mathrm{dx}$

$= {x}^{2} / 2 \arcsin {x}^{2} - \int {x}^{2} / 2 {\left(\arcsin {x}^{2}\right)}^{p} r i m e \mathrm{dx}$

Recognising that: $\frac{d}{\mathrm{dx}} \left(\arcsin \left(u\right)\right) = \frac{1}{\sqrt{1 - {u}^{2}}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

$= {x}^{2} / 2 \arcsin {x}^{2} - \int {x}^{2} / 2 \cdot \frac{2 x}{\sqrt{1 - {x}^{4}}} \mathrm{dx}$

$= {x}^{2} / 2 \arcsin {x}^{2} - \int {x}^{3} / \sqrt{1 - {x}^{4}} \mathrm{dx}$

$= {x}^{2} / 2 \arcsin {x}^{2} - \int 2 \cdot \left(- \frac{1}{4}\right) \cdot {\left(\sqrt{1 - {x}^{4}}\right)}^{p} r i m e \mathrm{dx}$

$= {x}^{2} / 2 \arcsin {x}^{2} + \frac{1}{2} \sqrt{1 - {x}^{4}} + C$

The definite integral is, therefore:

${\left[{x}^{2} / 2 \arcsin {x}^{2} + \frac{1}{2} \sqrt{1 - {x}^{4}}\right]}_{0}^{1}$

$= \left(\frac{1}{2} \cdot \frac{\pi}{2} + \frac{1}{2} \cdot 0\right) - \left(0 + \frac{1}{2}\right)$

$= \frac{\pi}{4} - \frac{1}{2}$