Calculus
posted by Shelley .
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 (1x^4))dx
I don't know how to solve this. Can I take U = root 1x^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 (1x^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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