Solve \int(x^3/\sqrt(1-x^2))dx using two methods?

a. Integration by parts
b. Trigonometric Substitution

(I solved part B, but you can answer it if you want. I'm interested in seeing if I might have messed up in steps anywhere.)

1 Answer
Mar 2, 2018

I=-x^2sqrt(1-x^2)-2/3(1-x^2)^(3/2)+C

Explanation:

We want to solve

I=intx^3/(sqrt(1-x^2))dx

Use integration by parts

intudv=uv-intvdu

Let u=x^2=>du=2xdx

And dv=x/sqrt(1-x^2)dx=>v=-sqrt(1-x^2)

I=-x^2sqrt(1-x^2)+intsqrt(1-x^2)2xdx

Make a substitution s=1-x^2=>ds=-2xdx

I=-x^2sqrt(1-x^2)-intsqrt(s)ds

=-x^2sqrt(1-x^2)-2/3s^(3/2)+C

Substitute back s=1-x^2

I=-x^2sqrt(1-x^2)-2/3(1-x^2)^(3/2)+C

Alternative method

This can also be done by ordinary substitution, we seek

I=intx^3/(sqrt(1-x^2))dx

Make a substitution u=1-x^2=>du=-2xdx

I=-1/2intx^2/(sqrt(u))du

But u=1-x^2=>x^2=1-u

I=-1/2int(1-u)/(sqrt(u))du=-1/2intu^(-1/2)-u^(1/2)du

Which is

I=-u^(1/2)+1/3u^(3/2)+C

Substitute back u=1-x^2

I=-(1-x^2)^(1/2)+1/3(1-x^2)^(3/2)+C