Question

How do you solve `intxsqrt(x^2+8)*dx` using trig substitution?

Answer 1

Let, `sqrt(x^2+8)=t=>x^2+8=t^2=>2xdx=2tdt=>xdx=tdt`
`:.I=intsqrt(x^2+8)*xdx=intt*tdt=intt^2dt=t^3/3+c`
`=>I=1/3(t)^3+c,where,t=sqrt(x^2+8)`
`=1/3(sqrt(x^2+8))^3+c`

Explanation:

Above answer is Without Trig.Substitution.

Answer below is With Trig.Substitution.

Decide which answer is better ?

Here,

`I=intxsqrt(x^2+8)dx`

Let, `x=2sqrt2tanu=>dx=2sqrt2sec^2udu`

`and tanu=x/(2sqrt2)=>tan^2u=x^2/8.`

`:.I=int2sqrt2tanusqrt(8tan^2u+8)*2sqrt2sec^2udu`

`=int2sqrt2tanu2sqrt2sqrt(tan^2u+1)*2sqrt2sec^2udu`

`=8inttanusecu*2sqrt2sec^2udu`

`=8(2sqrt2)int[secu]^2(secutanu)du`

`=16sqrt2int[secu]^2d/(dx)(secu)du`

`=16sqrt2([secu]^3/3)+c`

`=(16sqrt2)/3[sqrt(tan^2u+1)]^3+c,where,tan^2u=x^2/8`

`=(16sqrt2)/3[sqrt(x^2/8+1)]^3+c`

`=(16sqrt2)/3(sqrt(x^2+8)/(2sqrt2))^3+c`

`=(16sqrt2)/3xx(sqrt(x^2+8))^3/(16sqrt2)+c`

`=1/3(sqrt(x^2+8))^3+c`