Posted by **Shelley** on Monday, January 22, 2007 at 7:28pm.

Hello Everyone,

I need help with Calc II.

1. Integral from 0 to 1 of

(sin(3*pi*t))dt

For this one, I got -1/3pi cos (9 pi^2) + 1/3pi

2. indefinite integral of

sinxcos(cosx)dx

I got sin(cosx) + C

3. Indefinite integral of

x over (root (1-x^4))dx

I don't know how to solve this. Can I take U = root 1-x^4 or without the root?

Thank you for your help!

First is way wrong.

INT sin 3PIt= - 1/3PI * cos 3PI t

Second is right.

Third. I don't see it either. Will think on it, if I see a solution, I will post it later.

3. Indefinite integral of

x over (root (1-x^4))dx

Substitute x = sqrt[sin(t)]

You find that the indefinite integral in terms of t is:

1/2 t + c

So, in terms of x it is:

1/2 arcsin(x^2) + c

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