Question

How do you integrate `int 1/sqrt(4x^2+16x)` using trigonometric substitution?

Answer 1

Let `I=int1/sqrt(4x^2+16x)dx=1/2int1/sqrt(x^2+4x)dx`

Now, `sqrt(x^2+4x)=sqrt(x^2+4x+4-4)=sqrt{(x+2)^2-2^2}`.

We take substn. `"(x+2)=2secy, so, dx=2secytanydy"`

Also, `sqrt(x^2+4x)=sqrt{4sec^2y-4}=sqrt(4tan^2y)=2tany`.

Hence, `I=1/2int(2secytany)/(2tany) dy=1/2intsecydy`

`=1/2ln|secy+tany|`

`=1/2ln|(x+2)/2+sqrt(x^2+4x)/2|+C`

`=1/2ln|(x+2+sqrt(x^2+4x))/2|+C`

`=1/2ln|x+2+sqrt(x^2+4x)|+C-1/2ln2`

`=1/2ln|x+2+sqrt(x^2+4x)|+K, "where", K=C-1/2ln2`.

Enjoy Maths.!