How do you integrate #int x^2sqrt(16-x^2)# by trigonometric substitution?

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Answer 1 · 2018-03-23T07:46:56

The answer is #=32arcsin(x/4)-2sqrt(16-x^2)(x-1/8x^3)+C#

Perform the substitution

#x=4sintheta#, #=>#, #dx=4costhetad theta#

#sqrt(16-x^2)=sqrt(16-16sin^2theta)=4costheta#

Therefore,

The integral is

#I=intx^2sqrt(16-x^2)dx=int16sin^2theta*4costheta*4costheta d theta#

#=256intsin^2thetacos^2theta d theta#

#2sinthetacostheta=sin(2theta)#

#I=64intsin^2(2theta)d theta#

#cos(4theta)=1-2sin^2(2theta)#

#sin^2(theta)=(1-cos(4theta))/2#

Therefore,

#I=32int(1-cos(4theta))d theta#

#=32(theta-sin(4theta)/4)#

#=32arcsin(x/4)-8sin(4theta)#

#sin(4theta)=costheta(4sintheta-8sin^3theta)#

#=sqrt(16-x^2)/4(x-1/8x^3)#

And finally,

#intx^2sqrt(16-x^2)dx=32arcsin(x/4)-2sqrt(16-x^2)(x-1/8x^3)+C#