# Inverse Trig Integration

• Sep 10th 2008, 02:57 PM
veronicak5678
Inverse Trig Integration
I am trying to integrate arcsin (x^(1/2)).

I used a u-sub for x^(1/2) and tried to finish with integration by parts, but I ended up with something just as complicated, requiring another integration by parts. I think I'm going about it the wrong way...
• Sep 10th 2008, 03:39 PM
mr fantastic
Quote:

Originally Posted by veronicak5678
I am trying to integrate arcsin (x^(1/2)).

I used a u-sub for x^(1/2) and tried to finish with integration by parts, but I ended up with something just as complicated, requiring another integration by parts. I think I'm going about it the wrong way...

Showing your work would help. After integration by parts you should get an integral looking like $\displaystyle \int \frac{\sqrt{x}}{\sqrt{1-x}} \, dx$. Is this integral the cause of your trouble?
• Sep 10th 2008, 03:55 PM
veronicak5678
After integrating by parts, I end up with

u^2 sin ^-1 [u] - int ((u^2) / (1-u^2)^(1/2))
• Sep 10th 2008, 07:44 PM
mr fantastic
Quote:

Originally Posted by veronicak5678
After integrating by parts, I end up with

u^2 sin ^-1 [u] - int ((u^2) / (1-u^2)^(1/2))

Integration by parts formula: $\displaystyle \int u \, dv = uv - \int v \, du$.

$\displaystyle I = \int \arcsin (\sqrt{x}) \, dx$.

Make the following choice:

$\displaystyle u = \arcsin (\sqrt{x}) \Rightarrow du = - \frac{1}{2 \sqrt{x} \, \sqrt{1 - x}}$.

$\displaystyle dv = dx \Rightarrow v = x$.

Then $\displaystyle I = x \, \arcsin (\sqrt{x}) - \frac{1}{2} \int \frac{\sqrt{x}}{\sqrt{1 - x}} \, dx$.

Now refer to post #2.