How do you find the antiderivative of #int x^2sqrt(1-x^2) dx#?
1 Answer
Explanation:
#I=intx^2sqrt(1-x^2)dx#
Let
#I=intsin^2(theta)sqrt(1-sin^2(theta))(cos(theta)d theta)#
#I=intsin^2(theta)cos^2(theta)d theta#
Now let
#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
#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
#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
#I=1/8sin^-1(x)-1/16sin(2theta)cos(2theta)#
Reusing the sine double-angle identity and also applying
#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
#I=1/8sin^-1(x)+1/8sin(theta)sqrt(1-sin^2(theta))(2sin^2(theta)-1)#
Finally using
#I=1/8sin^-1(x)+1/8x(2x^2-1)sqrt(1-x^2)+C#