How do you integrate #int x^3 /sqrt(16 - x^2) dx# using trigonometric substitution?

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Answer 1 · 2016-03-05T15:43:42

The answer is: #I=-1/3sqrt(16-x^2)*(x^2+32)+c#.

If we make this substitution:

#x=4sintrArrdx=4cost*dt#,

then:

#int(64sin^3t)/sqrt(16-16sin^2t)*4cost*dt=#

#=int(64sin^3t)/(4sqrt(1-sin^2t))*4cost*dt=#

#=int(64sin^3t)/(4cost)*4cost*dt=64intsin^3t*dt=#

#64intsintsin^2t*dt=64intsint(1-cos^2t)dt=#

#64int(sint-sintcos^2t)dt=64(-cost+cos^3t/3)+c=(1)#.

Now, remembering our substitution:

#sint=x/4rArrcost=sqrt(1-sin^2)=sqrt(1-x^2/16)=1/4sqrt(16-x^2)#.

So:

#(1)=64(-1/4sqrt(16-x^2)+1/64*1/3*sqrt(16-x^2)*(16-x^2))+c=#

#=-16sqrt(16-x^2)+1/3sqrt(16-x^2)(16-x^2)+c=#

#=1/3sqrt(16-x^2)(-48+16-x^2)+c=#

#=-1/3sqrt(16-x^2)*(x^2+32)+c#.