Question

How do you integrate `int 1/sqrt(4x^2+16x+8)` using trigonometric substitution?

Answer 1

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

Explanation:

It is in general a good idea to divide any polynomial that we have to work with through by its leading coefficient; this almost always makes it simpler to work with. So:

`int1/sqrt(4x^2+16x+8)dx=1/2int1/sqrt(x^2+4x+2)dx`

Complete the square:

`1/2int1/sqrt((x+2)^2-2)dx`

Make a substitution to cause the square root to vanish via a trig identity. In this case we'll use `x+2=sqrt(2)sectheta`. We could also use `csc` (which would be similar) or, straying into hyperbolic functions, `cosh` (which would be easier).

As `x=-2+sqrt(2)sectheta`, `dx/(d theta)=sqrt(2)secthetatantheta`, and so our integral becomes

`1/2int1/sqrt(2sec^2theta-2)*sqrt(2)secthetatanthetad theta`

Recalling the identity `tan^2theta=sec^2theta-1`, we obtain

`1/2int1/(sqrt(2)tantheta)*sqrt(2)secthetatanthetad theta`

which simplifies considerably to become

`1/2intsectheta d theta`

The integral of `secx` is `ln|secx+tanx|`. Two demonstrations of this are given here: https://www.math.ubc.ca/~feldman/m121/secx.pdf
So we get

`1/2ln|sectheta+tantheta|+C`

Transform back to our original variable, `x`:

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

Rearrange a little to obtain a simpler form:

`1/2ln|(x+2)/sqrt(2)+sqrt(((x+2)^2-2)/2)|+C`
`1/2ln|x+2+sqrt((x+2)^2-2)|-1/2lnsqrt(2)+C`

Note that the `-1/2lnsqrt(2)` term is a constant and can therefore be folded into the integration constant `C`

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

If we wished to write this in the same style as the original question formula, we could use the same observation about the integration constant to add `1/2ln2` without changing the answer:

`1/2ln|x+2+sqrt(x^2+4x+2)|+C`
`1/2ln|x+2+sqrt(x^2+4x+2)|+1/2ln2+C`
`1/2ln|2(x+2)+sqrt(4x^2+16x+8)|+C`

It's arguable whether it's better to write it so that the expression inside the root matches that in the original integral or so that it is a simpler expression. I think I prefer our first answer myself:

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

It's more elegant.

Answer 2

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

Explanation:

Here,

`I=int1/sqrt(4x^2+16x+8)dx`

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

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

`I=1/2int1/sqrt((x+2)^2-(sqrt2)^2)dx`

Subst. `x+2=sqrt2secu=>dx=sqrt2secutanudu`

`=>secu=(x+2)/sqrt2`

`I=1/2int1/sqrt(2sec^2u-2)xxsqrt2secutanudu`

`:.I=1/2int(sqrt2secutanu)/(sqrt2tanu)du`

`=>=1/2intsecudu`

`:.I=1/2ln|secu+tanu|+c`

`=1/2ln|secu+sqrt(sec^2u-1)|+c`

Subst. back , `secu=(x+2)/sqrt2`

`I=1/2ln|(x+2)/sqrt2+sqrt((x+2)^2/2-1)|+c`

`I=1/2ln|(x+2)/sqrt2+sqrt((x+2)^2-2)/sqrt2|+c`

`I=1/2ln|((x+2)+sqrt(x^2+4x+2))/sqrt2|+c`

`I=1/2ln|(x+2)+sqrt(x^2+4x+2)|-1/2lnsqrt2+c`

`I=1/2ln|(x+2)+sqrt(x^2+4x+2)|+C,`
# where,C=c-1/2lnsqrt2#