Monday

April 21, 2014

April 21, 2014

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

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 dont 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

**Related Questions**

trig - Solve cos x-1 = sin^2 x Find all solutions on the interval [0,2pi) a. x=...

pre-calc - Can someone help me find the exact value of 4csc(3pi/4)-cot(-pi/4)? ...

Pre-Calculus-check answers - State the period and phase shift of the function y...

Linear Algebra - Hello, I'm trying to find the Fourier Series of a function ...

Linear Algebra - Hello, I'm trying to find the Fourier Series of a function ...

CALCULUS LIMITS - What is the following limit? lim as n goes to infinity of (pi/...

trigonometry - 1) Perform the operation and leave the result in trig. form. [3/4...

math - u=(2 cos pi/4)i + (2 sin pi/4)j, v = (cos 3pi/2)i + (sin 3pi/2)j find the...

Math - Test for symmetry with repsect to Q=pi/2, the polar axis, and the pole. r...

Math - Test for symmetry with repsect to Q=pi/2, the polar axis, and the pole. ...