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

1 Answer
Apr 18, 2018

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