Question

How do you evaluate the integral `sqrt(36+9x^2)dx`?

Answer 1

Factor, then use trignometric or hyperbolic trigonometric substitution.

`sqrt(36+9x^2)=3sqrt(4+x^2)`

We can find `intsqrt(4+x^2)dx` by substituting `x=2tantheta` so `dx=2sec^2thetad theta` and `sqrt(4+x^2)=2sectheta`

You then need to evaluate `2intsec^3theta d theta`, which is a but of work.

If you have hyperbolic trigonometric functions available, then there is a 'cleaner' solution.

Let `x=2sinht` which makes `dx=2coshtdt` and `sqrt(4+x^2)=2cosht`

Now you need to evaluate `4intcosh^2tdt=2int(cosh2t+1)dt`. this is not difficult, then back-substitute.