Integrate x*sqrt(1-x^4) from 0 to 1

I'm asked to make a substitution and then evaluate the resulting integral in terms of area. I've tried letting u = 1-x^4 and letting u = x^2 both to no avail. I'm lost.

Any help would be appreciated!

Your second substitution which was u = x^2 seems correct.

∫x√(1 - x^4) dx from 0 to 1

u = x^2

du/2 = x dx

1/2*∫√(1 - u^2) from 0 to 1

Now note that the integral represents the area of a quarter circle of radius 1. Therefore, the answer is:

(1/2)(pi/4) =pi/8

This question relies on knowing or looking up the standard integral
∫1/√(1 - u^2) du = sin^-1(u)
You correctly guessed substitution u = x^2

A common mistake here is to just replace x^2 by u but neglect the dx part
du/dx = 2x so 2x dx = du That helps a lot because we can rewrite
∫x/√(1 - x^4) dx = (1/2)∫1/√(1 - x^4)*2x dx = (1/2)∫1/√(1 - u^2) du
= (1/2)sin^-1(u) = (1/2)sin^-1(x^2)

Regards - Ian

suppose e try u=x^2
then we differentiate it and get du=2xdx

or

du/2=xdx

then your integral is of

0.5*sqrt(1-u^2)du

note: integration limits remain same