1. ## Definite Integrals!

$\displaystyle \int^1_0 arcsin\left(\frac{2x}{1+x^2}\right)$

PS. How do you get the brackets to enclose the entire expression?

2. Originally Posted by polymerase
PS. How do you get the brackets to enclose the entire expression?
well, i can answer that part

type "\left(" and "\right)"

3. Originally Posted by polymerase
$\displaystyle \int^1_0 arcsin\left(\frac{2x}{1+x^2}\right)$
I was tryin' to find a nice way to tackle this but I couldn't.

Looks like it's pretty straightforward.

Integrate by parts:

$\displaystyle u = \arcsin \frac{{2x}} {{1 + x^2 }} \implies du = \frac{2} {{1 + x^2 }}\,dx$ & $\displaystyle dv=dx,$ so

$\displaystyle \int_0^1 {\arcsin \frac{{2x}} {{1 + x^2 }}\,dx} = \left. {x\arcsin \frac{{2x}} {{1 + x^2 }}} \right|_0^1 - \int_0^1 {\frac{{2x}} {{1 + x^2 }}\,dx} .$

The rest follows.

4. Originally Posted by Krizalid
Integrate by parts:
that's what i thought...but i was too lazy to check...

5. Originally Posted by Krizalid
I was tryin' to find a nice way to tackle this but I couldn't.
I know you dont like this, but you could use the substitution $\displaystyle u=tan\;x$ then use by parts on $\displaystyle \int 2u\;sec^2\;u\;du$

6. Anyway you can't get rid of integration by parts.