Integration by Substitution

∫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! :)

1 Attachment(s)

Re: Integration by Substitution

Re: Integration by Substitution

Re: Integration by Substitution

Hello, Chaim!

We have: .

Let

Substitute: .

Back-substitute: .

Re: Integration by Substitution

Thanks for responding guys!

I understand how u and du is coming from now.

I'm a bit confused on one thing.

The *xdx*

Just wondering, how come *xdx* = -1/2?

Re: Integration by Substitution

It isn't and Soroban didn't say it was. From , we get so that . Divide both sides by -2 to get .