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

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

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Answer 1 · 2015-03-06T03:14:08

Factor, then use trignometric or hyperbolic trigonometric substitution.

$\sqrt{36 + 9 x^{2}} = 3 \sqrt{4 + x^{2}}$ $sqrt(36+9x^2)=3sqrt(4+x^2)$

We can find $\int \sqrt{4 + x^{2}} d x$ $intsqrt(4+x^2)dx$ by substituting $x = 2 tan θ$ $x=2tantheta$ so $d x = 2 {sec}^{2} θ d θ$ $dx=2sec^2thetad theta$ and $\sqrt{4 + x^{2}} = 2 sec θ$ $sqrt(4+x^2)=2sectheta$

You then need to evaluate $2 \int {sec}^{3} θ d θ$ $2intsec^3theta d theta$ , which is a but of work.

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

Let $x = 2 sinh t$ $x=2sinht$ which makes $d x = 2 cosh t d t$ $dx=2coshtdt$ and $\sqrt{4 + x^{2}} = 2 cosh t$ $sqrt(4+x^2)=2cosht$

Now you need to evaluate $4 \int {cosh}^{2} t d t = 2 \int (cosh 2 t + 1) d t$ $4intcosh^2tdt=2int(cosh2t+1)dt$ . this is not difficult, then back-substitute.

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

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