∫5x

^{3}√(1-x

^{2})

*dx*
So I'm still a beginner with integrating by substituting.

What I know is that you use 'u' from one of the thing in the equation to substitute.

Though, I'm not entirely sure where 'u' should be coming from.

So I did

__u = (1-x__^{2})
And since it has a cube root, I did this:

1/3

*du* =

^{3}√(1-x

^{2})

*dx*
So I got: ∫1/3 u

^{1/3} *du*

4/3 u

^{4/3} + C -> 4/3 (1-x

^{2})

^{3} + C

But I'm not sure if this is entirely correct after:

Someone else showed me how to do the steps, which was this:

∫5x

^{3}√(1-x

^{2})

*dx* = -5/2∫(1-x

^{2})

^{1/3}(-2x)

*dx* = -5/2 * ((1-x

^{2})

^{4/3})/(4/3) + C =

__(-15/8) * (1-x__^{2})^{4/3} + C
This was the correct answer in the book.

**But I'm a bit confused on how the person got the -5/2, and the (-2x)**
Can someone explain to me that part?

Thanks!