Question

How do you integrate `int 1/sqrt(4x+8sqrtx+12)` using trigonometric substitution?

Answer 1

`sqrt(x+2sqrtx+3)-ln(abs(sqrtx+sqrt(x+2sqrtx+3)+1))+C`

Explanation:

We have:

`intdx/sqrt(4x+8sqrtx+12)`

Factor `sqrt4=2` from the denominator:

`=1/2intdx/sqrt(x+2sqrtx+3)`

Complete the square in the denominator. Think about it in terms of `x^2+2x+3`, and then switch `x` terms to `sqrtx` and `x^2` to `x`:

`=1/2intdx/sqrt((sqrtx+1)^2+2)`

Now, let `sqrtx+1=sqrt2tantheta`. Note that this implies that `dx/(2sqrtx)=sqrt2sec^2thetad theta`.

Rearranging some:

`=int(sqrtxdx)/(2sqrtxsqrt((sqrtx+1)^2+2))`

Now, performing the substitutions. We have `dx/(2sqrtx)=sqrt2sec^2thetad theta` present, as well as `sqrtx=sqrt2tantheta-1` in the numerator. Don't forget the switch in the radical in the denominator either:

`=int((sqrt2tantheta-1)sqrt2sec^2thetad theta)/sqrt(2tan^2theta+2)`

Factoring `sqrt2` from the denominator and using `sqrt(tan^2theta+1)=sectheta`:

`=int((sqrt2tantheta-1)sqrt2sec^2theta)/(sqrt2sectheta)d theta`

`=int(sqrt2tantheta-1)secthetad theta`

`=sqrt2inttanthetasecthetad theta-intsecthetad theta`

Both of which are common integrals:

`=sqrt2sectheta-ln(abs(sectheta+tantheta))`

Write `sectheta` in terms of `tantheta`:

`=sqrt2sqrt(tan^2theta+1)-ln(abs(sqrt(tan^2theta+1)+tantheta))`

Using `tantheta=(sqrtx+1)/sqrt2`:

`=sqrt2sqrt((sqrtx+1)^2/2+1)-ln(abs(sqrt((sqrtx+1)^2/2+1)+(sqrtx+1)/sqrt2))`

`=sqrt2sqrt(((sqrtx+1)^2+2)/2)-ln(abs(sqrt(((sqrtx+1)^2+2)/2)+(sqrtx+1)/sqrt2))`

`=sqrt((sqrtx+1)^2+2)-ln(abs((sqrt((sqrtx+1)^2+2)+sqrtx+1)/sqrt2))`

Note that the `1/sqrt2` can actually be taken from the denominator of the logarithm as `-lnsqrt2`, which will be absorbed into the constant of integration in the next step:

`=sqrt(x+2sqrtx+3)-ln(abs(sqrtx+sqrt(x+2sqrtx+3)+1))+C`