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

1 Answer
maganbhai P.
Apr 7, 2018

Sometimes trig.substitution is not easy. See and compare.
Let,#16-x^2=t^2=>xdx=-tdt#
#I=-int(16-t^2)t^2dt=int(t^4-16t^2)dt=t^5/5-(16t^3)/3+c#
#I=(sqrt(16-x^2))^5/5-(16(sqrt(16-x^2))^3)/3+C#

Explanation:

Here,

#I=intx^3sqrt(16-x^2)dx=intx^2sqrt(16-x^2)*xdx#

Let, #x=4sinu=>x^2=16sin^2u=>2xdx=32sinucosudu#

i.e. #xdx=16sinucosudu#

#=>I=int(16sin^2u)sqrt(16-16sin^2u)(16sinucosu)du#

#I=16xx4xx16intsin^2u(cosu)sinucosudu#

#I=1024intsin^2ucos^2usinu du#

#=-1024int(1-cos^2u)cos^2u(-sinu)du#

#=-1024[int(cosu)^2(-sinu)du-int(cosu)^4(-sinu)du#

#=-1024[(cosu)^3/3-(cosu)^5/5]+Ctowhere,sin^2u=x^2/16#

#=-1024[(sqrt(1-sin^2u))^3/3-(sqrt(1-sin^2u))^5/5]+C#

#=-1024[(sqrt(1-x^2/16))^3/3-(sqrt(1-x^2/16))^5/5]+C#

#=-1024[(sqrt(16-x^2))^3/(3xx64)-(sqrt(16- x^2))^5/(5xx1024)]+C#

#=-[(16(sqrt(16-x^2))^3)/3-(sqrt(16-x^2))^5/5]+C#

#I=(sqrt(16-x^2))^5/5-(16(sqrt(16-x^2))^3)/3+C#

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