Question

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.)

Answer 1

`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`