∫(sin⁻¹x)²

**raj007** · June 25th 2010, 08:55 AM

$∫(sin⁻¹x)²$

the only problem is i have to do it via method of substitution, i dont have to to use integration by parts

**Also sprach Zarathustra** · June 25th 2010, 09:27 AM

Let use integration by parts.

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

Which gives us:

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

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

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

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

So,

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

**Also sprach Zarathustra** · June 25th 2010, 10:53 AM

By substitution:

let $t=arcsinx$

so, $sint=x$

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

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

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

I let you finishing this (by parts)

I guess there is no escape from integration by parts...

**Also sprach Zarathustra** · June 25th 2010, 10:53 AM

By substitution:

let $t=arcsinx$

so, $sint=x$

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

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

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

I let you finishing this (by parts)

I guess there is no escape from integration by parts...

**raj007** · June 25th 2010, 07:32 PM

i guess the same, i hope the question is misprinted in the substitution exercise,

thanks everyone for their time.

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