# ∫(sin⁻¹x)²

• Jun 25th 2010, 08:55 AM
raj007
∫(sin⁻¹x)²
$\displaystyle ∫(sin⁻¹x)²$

the only problem is i have to do it via method of substitution, i dont have to to use integration by parts
• Jun 25th 2010, 09:27 AM
Also sprach Zarathustra
Let use integration by parts.

Let $\displaystyle u = [ arcsin(x) ]^2; dv = dx; du = 2arcsin(x) (\frac{1}{\sqrt(1 - x^2)}) dx; v = x.$

Which gives us:

$\displaystyle x [arcsin(x)]^2 - \int ( 2x arcsin(x) (\frac{1}{\sqrt(1 - x^2)}) )$
Using integration by parts again.

Let $\displaystyle u = arcsin(x); dv = \frac{2x}{\sqrt(1 - x^2)} dx. du = \frac{1}{\sqrt(1 - x^2)} dx; v = (-1)\sqrt(1 - x^2)$
$\displaystyle x [arcsin(x)]^2 - [ (-1) arcsin(x) \sqrt(1 - x^2) - \int ( [\frac{1}{\sqrt(1 - x^2)}] [(-1)\sqrt{(1 - x^2)^(1/2)}] du )$

$\displaystyle x [arcsin(x)]^2 + arcsin(x) \sqrt{(1 - x^2)} + \int ( [\frac{1}{\sqrt{(1 - x^2)}}] [(-1)\sqrt{(1 - x^2)^(1/2)}] du )$
$\displaystyle x [arcsin(x)]^2 + arcsin(x) \sqrt{(1 - x^2)} - \int ( [\frac{1}{\sqrt{(1 - x^2)}}] [\sqrt{(1 - x^2)}] du )$

$\displaystyle x [arcsin(x)]^2 + arcsin(x) \sqrt{(1 - x^2)} - \int dx$

So,

$\displaystyle x [arcsin(x)]^2 + arcsin(x) \sqrt{(1 - x^2)} - x + C$
• Jun 25th 2010, 10:53 AM
Also sprach Zarathustra
By substitution:

let $\displaystyle t=arcsinx$

so, $\displaystyle sint=x$

$\displaystyle dt=\frac{dx}{\sqrt{1-x^2}$

$\displaystyle dx=\sqrt{1-sin^2t}dt=costdt$

$\displaystyle \int (arcsinx)^2dx=\int t^2costdt$

I let you finishing this (by parts)

I guess there is no escape from integration by parts...
• Jun 25th 2010, 10:53 AM
Also sprach Zarathustra
...
By substitution:

let $\displaystyle t=arcsinx$

so, $\displaystyle sint=x$

$\displaystyle dt=\frac{dx}{\sqrt{1-x^2}}$

$\displaystyle dx=\sqrt{1-sin^2t}dt=costdt$

$\displaystyle \int (arcsinx)^2dx=\int t^2costdt$

I let you finishing this (by parts)

I guess there is no escape from integration by parts...
• Jun 25th 2010, 07:32 PM
raj007
i guess the same, i hope the question is misprinted in the substitution exercise,

thanks everyone for their time.