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
Øko
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#

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