Question

How do you integrate `int 1/(x-sqrt(9 + x^2))dx` using trigonometric substitution?

Answer 1

`-(x^2+xsqrt(x^2+9)+9lnabs(sqrt(x^2+9)+x))/18+C`

Explanation:

First multiply by the conjugate:

`I=int1/(x-sqrt(9+x^2))*(x+sqrt(9+x^2))/(x+sqrt(9+x^2))dx`

`=int(x+sqrt(9+x^2))/(x^2-(9+x^2))dx`

`=-1/9int(x+sqrt(9+x^2))dx`

`=-1/9intxdx-1/9intsqrt(9+x^2)dx`

`=-1/18x^2-1/9intsqrt(9+x^2)dx`

Letting `J=intsqrt(9+x^2)dx`, we will solve this by using the substitution `x=3tantheta`. This implies that `dx=3sec^2thetad theta`.

`J=intsqrt(9+x^2)dx=intsqrt(9+9tan^2theta)(3sec^2thetad theta)`

`=9intsqrt(1+tan^2theta)(sec^2thetad theta)`

Since `1+tan^2theta=sec^2theta`:

`J=9intsec^3thetad theta`

To solve this integral, we will use integration by parts, which takes the form `intudv=uv-intvdu`. For `K=intsec^3thetad theta`, let:

`{(u=sectheta" "=>" "du=secthetatanthetad theta),(dv=sec^2thetad theta" "=>" "v=tantheta):}`

Thus:

`K=secthetatantheta-intsecthetatan^2thetad theta`

Let `tan^2theta=sec^2theta-1`:

`K=secthetatantheta-intsectheta(sec^2theta-1)d theta`

`K=secthetatantheta-intsec^3thetad theta+intsecthetad theta`

Notice that we have the original integral `K=intsec^3thetad theta` imbedded in this integral. Also, the integration of just secant is a common integral.

`K=secthetatantheta-K+lnabs(sectheta+tantheta)`

Solving for `K`:

`2K=secthetatantheta+lnabs(sectheta+tantheta)`

`K=(secthetatantheta+lnabs(sectheta+tantheta))/2=intsec^3thetad theta`

Returning to `J=9intsec^3thetad theta`, we see that:

`J=9/2(secthetatantheta+lnabs(sectheta+tantheta))`

We should write this in terms of just `tantheta`:

`J=9/2(sqrt(tan^2theta+1)(tantheta)+lnabs(sqrt(tan^2theta+1)+tantheta))`

Since `tantheta=x/3`:

`J=9/2(sqrt(x^2/9+1)(x/3)+lnabs(sqrt(x^2/9+1)+x/3))`

`J=9/2(sqrt(1/9(x^2+9))(x/3)+lnabs(sqrt(1/9(x^2+9))+x/3))`

`J=9/2(x/9sqrt(x^2+9)+lnabs(1/3(sqrt(x^2+9)+x)))`

`J=x/2sqrt(x^2+9)+9/2lnabs(sqrt(x^2+9)+x)-9/2ln(3)`

(Ignore the constant since this is from an integral, just we haven't yet added `C` for convenience purposes.)

`intsqrt(9+x^2)dx=x/2sqrt(x^2+9)+9/2lnabs(sqrt(x^2+9)+x)`

Plugging this into `I=-1/18x^2-1/9intsqrt(9+x^2)dx`:

`I=-1/18x^2-1/9(x/2sqrt(x^2+9)+9/2lnabs(sqrt(x^2+9)+x))`

`I=-1/18x^2-1/18xsqrt(x^2+9)-1/2lnabs(sqrt(x^2+9)+x)`

`I=-(x^2+xsqrt(x^2+9)+9lnabs(sqrt(x^2+9)+x))/18+C`