Question

Integral Question

Answer 1

I really just need help. I don't think it involves a crazy trig substitution but a U substitution.

Answer 2

\[4\int\limits_{0}^{1}x^3 \sqrt{1-x^2}dx\]

Answer 3

Like let U=1-x^2
DU= -2x dx???

Answer 4

But you'll have x^3 left there.

Answer 5

I suppose this can be made to work...

Answer 6

I just want the easiest way. If I'm wrong, its ok.

Answer 7

Have you tried 'integration by parts' to decrease the power of x^3 to 1

Answer 8

Ok... I'm going to remove the limits for now....
\[\Large 4\int x^3 \sqrt{1-x^2}dx\]

Answer 9

And you said
\[\large u = 1-x^2\\\large du = -2x dx\]

Answer 10

Thats what I first thought but what about integration by parts? I think I made this problem harder than it is.

Answer 11

Internet connection going in and out sorry If I don't respond

Answer 12

Well...
\[\Large -2\int x^2\sqrt{1-x^2} \ (-2x)dx\]

Answer 13

You follow?

Answer 14

Well, anyway, I need to go in a bit, I'll put the next step and the rest should be up to you:
\[\Large \Large -2\int x^2\sqrt{1-x^2} \ \color{red}{ (-2x)dx}\\\Large -2\int x^2\sqrt{\color{red}{1-x^2}} \ \color{blue}{du}\\\Large -2\int \color{red}{x^2}\sqrt{\color{blue}{u}} \ du\\\Large -2\int \color{blue}{(1-u)}\sqrt{u} \ du\]

then it's all power rule.