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

1 Answer
Mar 6, 2015

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.