Calculus

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