Inverse Trig Integration

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September 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...
September 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 $\int \frac{\sqrt{x}}{\sqrt{1-x}} \, dx$ . Is this integral the cause of your trouble?
September 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))
September 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: $\int u \, dv = uv - \int v \, du$ .

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

Make the following choice:

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

$dv = dx \Rightarrow v = x$ .

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

Now refer to post #2.