How do you find the antiderivative of #int x^2sqrt(1-x^2) dx#?

1 Answer
Feb 27, 2017

#1/8sin^-1(x)+1/8x(2x^2-1)sqrt(1-x^2)+C#

Explanation:

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

Let #x=sin(theta)#. This implies that #dx=cos(theta)d theta#. Then:

#I=intsin^2(theta)sqrt(1-sin^2(theta))(cos(theta)d theta)#

#I=intsin^2(theta)cos^2(theta)d theta#

Now let #cos^2(theta)=1-sin^2(theta)#:

#I=intsin^2(theta)(1-sin^2(theta))d theta#

#I=intsin^2(theta)d theta-intsin^4(theta)d theta#

Now use the identity #sin^2(alpha)=1/2(1-cos(2alpha))#:

#I=1/2int(1-cos(2theta))d theta-int(1/2(1-cos(2theta)))^2d theta#

#I=1/2intd theta-1/2intcos(2theta)d theta-1/4int(1-2cos(2theta)+cos^2(2theta))d theta#

Use the identity #cos^2(alpha)=1/2(1+cos(2alpha))#:

#I=1/2intd theta-1/2intcos(2theta)d theta-1/4intd theta+1/2intcos(2theta)d theta-1/8int(1+cos(4theta))d theta#

#I=1/4intd theta-1/8intd theta-1/8intcos(4theta)d theta#

Solving through a substitution:

#I=1/8theta-1/32sin(4theta)#

We now have to return to #x# from #theta#. Recall that #x=sin(theta)#. Thus #theta=sin^-1(x)#. Also, use the identity #sin(2alpha)=2sin(alpha)cos(alpha)# to break down #sin(4theta)#:

#I=1/8sin^-1(x)-1/16sin(2theta)cos(2theta)#

Reusing the sine double-angle identity and also applying #cos(2alpha)=1-2sin^2(alpha)#:

#I=1/8sin^-1(x)-1/8sin(theta)cos(theta)(1-2sin^2(theta))#

Flipping the order of the terms in the last parenthesis with the preceding negative sign and using #cos(x)=sqrt(1-sin^2(x))#:

#I=1/8sin^-1(x)+1/8sin(theta)sqrt(1-sin^2(theta))(2sin^2(theta)-1)#

Finally using #x=sin(theta)#:

#I=1/8sin^-1(x)+1/8x(2x^2-1)sqrt(1-x^2)+C#