Question

How do you integrate `int (6x^3)/sqrt(9+x^2) dx` using trigonometric substitution?

Answer 1

`2(x^2-18)sqrt(x^2+9)+C`.

Explanation:

Suppose that, `I=int(6x^3)/sqrt(9+x^2)dx=int(6x^3)/sqrt(x^2+3^2)dx`.

We subst. `x=3tanu. :. dx=3sec^2udu`.

`:. I=int{(6*27tan^3u)(3sec^2u)}/sqrt(9tan^2u+9)du`,

`=162int(tan^3usecu)`,

`162int(tan^2u*tanusecu)du`.

`=162int(sec^2u-1)secutanudu`.

If we take `secu=v," then, "secutanudu=dv`.

`:. I=162int(v^2-1)dv`,

`=162(v^3/3-v)`,

`=162/3v(v^2-3)`,

`=54secu(sec^2u-3)............[because, v=secu]`,

`=54sqrt(tan^2u+1){(tan^2u+1)-3}`,

`=54{sqrt((x/3)^2+1)}{(x/3)^2-2}`,

`rArr I=2(x^2-18)sqrt(x^2+9)+C`, as desired!

Feel the Joy of Maths.!

Answer 2

Here is a `2^(nd)` Method ( without using trgo. substn.) :

`2(x^2-18)sqrt(x^2+9)+C`

Explanation:

Let, `I=int(6x^3)/sqrt(x^2+9)dx=int{(3x^2)(2xdx)}/sqrt(x^2+9)`.

Subst. `x^2+9=t^2. :. x^2=t^2-9. :. 2xdx=2tdt`.

`:. I=int{3(t^2-9)(2tdt)}/sqrt(t^2)`,

`=6int(t^2-9)dt`,

`=6(t^3/3-9t)`,

`=2t(t^2-27)`,

`=2{(x^2+9)-27}sqrt(x^2+9)`.

`rArr I=2(x^2-18)sqrt(x^2+9)+C`, as in the `1^(st)` Method!