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.