How do you evaluate the integral sqrt(36+9x^2)dx36+9x2dx?

1 Answer
Mar 6, 2015

Factor, then use trignometric or hyperbolic trigonometric substitution.

sqrt(36+9x^2)=3sqrt(4+x^2)36+9x2=34+x2

We can find intsqrt(4+x^2)dx4+x2dx by substituting x=2tanthetax=2tanθ so dx=2sec^2thetad thetadx=2sec2θdθ and sqrt(4+x^2)=2sectheta4+x2=2secθ

You then need to evaluate 2intsec^3theta d theta2sec3θdθ, which is a but of work.

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

Let x=2sinhtx=2sinht which makes dx=2coshtdtdx=2coshtdt and sqrt(4+x^2)=2cosht4+x2=2cosht

Now you need to evaluate 4intcosh^2tdt=2int(cosh2t+1)dt4cosh2tdt=2(cosh2t+1)dt. this is not difficult, then back-substitute.