Question

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

Answer 1

`sqrt(16+x^2)+16/sqrt(16+x^2)`

Explanation:

Remember that
`tan^2 theta+1=sec^2 theta`

For the case is convenient that

`x=4tany`
`dx=4sec^2 y *dy`

Then

`int x^3/(sqrt(16+x^2))^3dx=int (64tan^3 y)/(sqrt(16+16tan^2 y))^3*4sec^2y* dy`
`=int (256 tan^3 y*sec^2 y)/(4sec y)^3dy=int (4 tan^3 y)/sec y dy`
`=4*int sin^3 y/cos^3 y*cos y*dy=4int sin y/cos^2 y*(1-cos^2 y) dy`
`=4int sin y/cos^2 y*dy-4int sin y*dy` [A]

`-> int sin y/cos^2 y*dy=`
Making `cos y=z` and `-sin y*dy=dz`
`=- int dz/z^2=1/z=1/cos y`

Inserting the result above in expression [A]:
`=4/cos y+4cos y +const.` [B]

But

`4tan y=x` => `siny=x/4*cos y`
`-> cos^2 y+sin^2 y = 1` => `cos^2 y+x^2/16*cos^2 y=1` => `(1+x^2/16)cos^2 y=1` => `cos y=sqrt(16/(16+x^2))=4/sqrt(16+x^2)`

Inserting the result above in expression [B]:

`=cancel(4)/(cancel(4)/sqrt(16+x^2))+4*4/sqrt(16+x^2)+const.`
`=sqrt(16+x^2)+16/sqrt(16+x^2)+const.`