Question

How do you integrate `int 1/sqrt(2x-12sqrtx-5)` using trigonometric substitution?

Answer 1

#I=sqrt(2x-12sqrtx-5)+3sqrt2ln|sqrt(2x)-3sqrt2+sqrt(2x-12sqrtx- 5)|+C,#
There is no any trig.function in the answer .So it is difficult to proceed with trig substitution.

Explanation:

Here,

`I=int1/sqrt(2x-12sqrtx-5)dx`

Let, `sqrtx=t=>x=t^2=>dx=2tdt`

`I=int(2t)/sqrt(2t^2-12t-5)dt`

`=1/2int(4t-12+12)/sqrt(2t^2-12t-5)dt`

#=color(red)(1/2int(4t-12)/sqrt(2t^2-12t-5)dt)+color(blue) (1/2int12/sqrt(2t^2-12t-5)dt#

`I=color(red)(I_1)+color(blue)(I_2)...to(A)`

Now, `I_1=1/2int(4t-12)/sqrt(2t^2-12t-5)dt`

Take, `sqrt(2t^2-12t-5)=u=>2t^2-12t-5=u^2`

`=>4t-12=2udu`

`:.I_1=1/2int(2u)/udu=intdu=u+c_1,tou=sqrt(2t^2-12t-5)`

`=sqrt(2t^2-12t-5)+c_1,where,sqrtx=t`

`:.I_1=sqrt(2x-12sqrtx-5)+c_1`

`Also, I_2=1/2int12/sqrt(2t^2-12t-5)dt`

`I_2=6/sqrt2int1/sqrt(t^2-6t-5/2)dt`

`=6/sqrt2int1/sqrt(t^2-6t+9-23/2)dtto[-5/2=(18-23)/2]`

`=3sqrt2int1/sqrt((t-3)^2-(sqrt(23/2))^2)dt`

`=3sqrt2ln|t-3+sqrt((t-3)^2-(sqrt(23/2))^2)|+c`

`=3sqrt2ln|t-3+color(violet)(sqrt(t^2-6t-5/2))|+c`

`=3sqrt2ln|t-3+sqrt(2t^2-12t-5)/color(orange)(sqrt2)|+color(orange)(c`

`I_2=3sqrt2ln|sqrt2t-3sqrt2+sqrt(2t^2-12t-5)|+color(orange)(c_2`

Subst .back `t=sqrtx`

`I_2=3sqrt2ln|sqrt(2x)-3sqrt2+sqrt(2x-12sqrtx-5)|+c_2`

Hence, from `(A)`,

#I=sqrt(2x-12sqrtx-5)+3sqrt2ln|sqrt(2x)-3sqrt2+sqrt(2x-12sqrtx- 5)|+C,#

`where,C=c_1+c_2`