Question

How do you integrate `int x^3sqrt(4-9x^2)` by trigonometric substitution?

Answer 1

`I=1/1215[3sqrt(4-9x^2))^5-20(sqrt(4-9x^2))^3]+c`

Explanation:

Here,

`I=intx^3 sqrt(4-9x^2)dx`

let,

`3x=2sinu=>x=2/3sinu=>dx=2/3cosudu and`

`9x^2=4sin^2u andsinu=(3x)/2=>cos^2u=1-(9x^2)/4`

`=>cos^2u=(4-9x^2)/4=>cosu=1/2sqrt(4-9x^2)`

So,

`I=int(2/3sinu)^3sqrt(4-4sin^2u)xx2/3cosudu`

`=int8/27sin^3u(2cosu)2/3cosudu`

`=32/81intsin^3ucos^2udu`

`=32/81intsin^2ucos^2usinudu`

`=-32/81int[(1-cos^2u)cos^2u(-sinu)du`

`=-32/81[intcos^2u(-sinu)du-intcos^4u(-sinu)du]`

`=-32/81[int(cosu)^2d/(dx)(cosu)du-int(cosu)^4d/(dx)(cosu)du]`

`=-32/81[(cosu)^3/3-(cosu)^5/5]+c`

`=-32/(81xx15)[5(cosu)^3-3(cosu)^5]+c`

Subst. `cosu=1/2sqrt(4-9x^2`

#I=-32/1215[5xx(sqrt(4-9x^2))^3/(8)-3xx(sqrt(4- 9x^2))^5/(32)]+c#

`I=-1/1215[20(sqrt(4-9x^2))^3-3(sqrt(4-9x^2))^5]+c`

`I=1/1215[3sqrt(4-9x^2))^5-20(sqrt(4-9x^2))^3]+c`

Answer 2

`I=1/1215[3(sqrt(4-9x^2))^5-20(sqrt(4-9x^2))^3]+c`

Explanation:

Method Without trigonometric substitution.

`I=intx^3sqrt(4-9x^2)dx=intx^2sqrt(4-9x^2)*xdx`

Let,

`sqrt(4-9x^2)=u=>4-9x^2=u^2=>9x^2=4-u^2`

`=>x^2=1/9(4-u^2)=>2xdx=-2/9udu`

`=>xdx=-1/9uduand`

So,

`I=int1/9(4-u^2)(u)(-1/9u)du`

`=1/81int(u^2-4)u^2du`

`=1/81int(u^4-4u^2)du`

`=1/81[u^5/5-(4u^3)/3]+c`

`=1/(81xx15)[3u^5-20u^3]+c`

Subst. `u=sqrt(4-9x^2)`

`I=1/1215[3(sqrt(4-9x^2))^5-20(sqrt(4-9x^2))^3]+c`