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

2 Answers
Jun 12, 2018

#8sin^-1(x/4)-(xsqrt(16-x^2))/2+C#

Explanation:

#I=intx^2/sqrt(16-x^2)color(red)(dx)#

Let's use the substitution #x=4sintheta#. This substitution implies that couple important things:

  • #x^2=16sin^2theta#
  • #sqrt(16-x^2)=sqrt(16-16sin^2theta)=4sqrt(1-sin^2theta)=4costheta#
  • #dx=4costhetad theta#

Now, let's substitute these back into our integral:

#I=int(16sin^2theta)/(4costheta)(4costhetad theta)=16intsin^2thetad theta#

To solve this, I like to use the identity #sin^2theta=1/2(1-cos2theta)#. If you've never seen this before, remember that #cos2theta=cos^2theta-sin^2theta#, which implies that #cos2theta=1-2sin^2theta#, from which you can solve for #sin^2theta# to get the identity from before. Using the identity in the integral:

#I=16int1/2(1-cos2theta)d theta=8(theta-1/2sin2theta)+C#

We need to get our final answer back in terms of #x# instead of #theta#. Use #sin2theta=2sinthetacostheta#:

#I=8theta-8sinthetacostheta+C#

Recall we started with #x=4sintheta#. Thus:

  • #sintheta=x/4#
  • #costheta=sqrt(1-sin^2theta)=sqrt(1-x^2/16)=sqrt(16-x^2)/4#
  • #theta=sin^-1(x/4)#

And finally:

#I=8sin^-1(x/4)-8(x/4)(sqrt(16-x^2)/4)+C#

#I=8sin^-1(x/4)-(xsqrt(16-x^2))/2+C#

Jun 12, 2018

#int x^2/sqrt(16-x^2) dx =8arcsin(x/4) - (x sqrt(16-x^2))/2 +C#

Explanation:

Let #x=4sint # with #t in(-pi/2,pi/2)#, #dx = 4costdt#, then:

#int x^2/sqrt(16-x^2) dx = int ((4sint)^2(4costdt))/sqrt(16-16sin^2t)#

#int x^2/sqrt(16-x^2) dx = 16int (sin^2tcostdt)/sqrt(1-sin^2t)#

For #t in(-pi/2,pi/2)# the cosine is positive, so:

#sqrt(1-sin^2t) = sqrt(cos^2t) = cost#

then:

#int x^2/sqrt(16-x^2) dx = 16int (sin^2tcostdt)/cost = 16 int sin^2tdt#

Use now the trigonometric identity:

#2sin^2alpha = 1-cos(2alpha)#

to have:

#int x^2/sqrt(16-x^2) dx = 8int (1-cos2t)dt#

and based on the linearity of the integral:

#int x^2/sqrt(16-x^2) dx = 8int dt - 4 int cos2td(2t)#

#int x^2/sqrt(16-x^2) dx = 8t - 4 sin2t +C#

#int x^2/sqrt(16-x^2) dx = 8t - 8 sintcost +C#

and undoing the substitution:

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

#int x^2/sqrt(16-x^2) dx =8arcsin(x/4) - (x sqrt(16-x^2))/2 +C#