Question

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

Answer 1

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`