# Thread: arcsin int by parts problem

1. ## arcsin int by parts problem

$\displaystyle \int xarcsinx^2dx$ on int [0,1]

Should I set u=arcsinx^2 and dv=x ??

2. That is correct. You then need to use substitution on the second integral. Just ask if you get stuck.

$\displaystyle t = x^2$

$\displaystyle dt = 2x \, dx$

$\displaystyle \frac{1}{2} \int 2x \arcsin(x^2) \, dx$

$\displaystyle \frac{1}{2} \int \arcsin(t) \, dt$

now go with parts ... let $\displaystyle u = \arcsin(t)$ and $\displaystyle dv = dt$

4. u=arcsin(t) -> du=$\displaystyle \frac{1}{\sqrt{1-x^2}}$
dv=dt -> v=t

=> (t)arcsin(t) - $\displaystyle \frac{1}{2}\int \frac{t}{\sqrt{1-x^2}}du$
= $\displaystyle x^2arcsin(x^2) - \frac {x^2}{2}arcsin(u) + C$ ???

I'm getting lost here....

5. you are back-substituting too soon ... get the integral done completely in terms of t, then back substitute when finished.

$\displaystyle u = \arcsin(t)$

$\displaystyle du = \frac{1}{\sqrt{1 - t^2}}$

$\displaystyle dv = dt$

$\displaystyle v = t$

$\displaystyle \frac{1}{2} \int \arcsin(t) \, dt = \frac{t\arcsin(t)}{2} - \frac{1}{2}\int \frac{t}{\sqrt{1-t^2}} \, dt$

$\displaystyle \frac{1}{2} \int \arcsin(t) \, dt = \frac{t\arcsin(t)}{2} + \frac{1}{4}\int \frac{-2t}{\sqrt{1-t^2}} \, dt$

$\displaystyle \frac{1}{2} \int \arcsin(t) \, dt = \frac{t\arcsin(t)}{2} + \frac{\sqrt{1-t^2}}{2} + C$

finish by back substituting $\displaystyle x^2$ for $\displaystyle t$ ...

$\displaystyle \frac{x^2\arcsin(x^2) + \sqrt{1 - x^4}}{2} + C$

6. where do you get the + 1/4 from???

7. $\displaystyle \frac{1}{2} \int \arcsin(t) \, dt = \frac{t\arcsin(t)}{2} \textcolor{red}{-\frac{1}{2}}\int \frac{t}{\sqrt{1-t^2}} \, dt$

$\displaystyle \frac{1}{2} \int \arcsin(t) \, dt = \frac{t\arcsin(t)}{2} + \textcolor{red}{\frac{1}{4}}\int \frac{\textcolor{red}{-2}t}{\sqrt{1-t^2}} \, dt$

the -2 is needed for the substitution required to find the antiderivative of the last integral expression.