how do i solve .....0 to pi/4 ∫ x^3*cos(x^2)dx .........im assuming set u =x^2..then what?

Aw, cool! I love questions like this! I would also set u equal to x^2. Then, you would have the integration of x^3cosu(du/2x).

Next, substitute u for x by solving u=x^2 for x. You should get x=u^(1/2). So, you should now have the integration of u^(3/2)cosu(du/2u^(1/2)). Sorry, I don't know how to do all the notation on the keyboard. *Note: u^(1/2) is the square root of u.

Now, you can cancel out the u's and bring the constant to the front. So, you should have 1/2 times the integration of u(cosu)du. You can integrate this a little easier.

ewwwww maths!! I haven't done maths for ages.. and that is also way too complicated for me lol

Sorry :(

then you put the pencil down and walk away...just walk away..