Question

Integrate `x^3/sqrt((x^2+16))` dx using trig substitution?

Answer 1

`intx^3/sqrt(x^2+16)dx=1/3(sqrt(16+x^2)/x)^3-sqrt(16+x^2)/x+c`

Explanation:

Let `x=4tanu`, then `dx=4sec^2udu`

also `secu=sqrt(1+x^2/16)=sqrt(16+x^2)/x`

and `I=intx^3/sqrt(x^2+16)dx`

= `int(64tan^3x)/sqrt(16tan^2u+16)xx4sec^2udu`

= `int(64tan^3x)/(4secu)xx4sec^2udu`

= `int64tan^3usecudu`

= `64int(sec^2u-1)(secutanu)du`

Let `v=secu`. Then `dv=secutanudu`:

and `I=int(v^2-1)dv`

= `1/3v^3-v+C`

= `1/3sec^3u-secu+C`

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

Answer 2

`= 4/3 sqrt{x^2 +4}(x^2-8)`

Explanation:

Try `x=4 tan theta`. Then `x = 4 sec^2theta d theta`
So
`int x^3/sqrt{x^2+16} dx = int {64 tan^3 theta times 4 sec^2theta}/{4sectheta}d theta`
`= 64 int tan^3theta sec theta d theta = 64 int tan^2theta tantheta sec theta d theta`
`=64 int (sec^2theta -1) d(sec theta) = 64 ({sec^3theta}/3 -sec theta)`
`= 64/3 sec theta (sec^2theta-3)=64/3 sqrt{1+x^2/16}(1+x^2/16-3)`
`= 1/3 sqrt{x^2 +16}(x^2-32)`