How do you integrate #int ( x^2 / sqrt(4 - x^2) ) dx#?

1 Answer
Aug 12, 2018

#int(x^2)/sqrt(4-x^2)dx=2sin^(-1)(x/2)-x/2sqrt(4-x^2)+C#, #C in RR#

Explanation:

#I=int(x^2)/sqrt(4-x^2)dx#
Let #x=2sin(theta)#

#dx=2cos(theta)d theta#

So:

#I=int(4sin(theta)^2)/sqrt(4-4sin(theta)^2)*2cos(theta)d theta#

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

Because #1-sin(theta)^2=cos(theta)^2#,

#I=2int2sin(theta)^2d theta#

Because #2sin(theta)^2=1-cos(2theta)#

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

#=2theta-sin(2theta)#

#=2(theta-sin(theta)cos(theta))#

Finally, because #theta=sin^(-1)(x/2)#, and #cos(sin^(-1)(x))=sqrt(1-x^2)#

#I=2sin^(-1)(x/2)-xsqrt(1-x^2/4)#

#=2sin^(-1)(x/2)-x/2sqrt(4-x^2)+C#, #C in RR#

\0/ Here's our answer !