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

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Answer 1 · 2015-03-06T03:14:08

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.

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