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Evaluate $\displaystyle\int \dfrac{1}{x^2+9} \, dx$. I've only learned the normal way of solving integrals but it does not work. I haven't learned how to use trigonometry to solve these problem.

I know you have to rearrange it into the form ${[f(x)]² + 1}$ and then integrate.

Can someone point me some rules to solve these kinds of question?

My teacher expected that the prerequisite course taught this but I have not learned it yet.

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  • $\begingroup$ $x=3t$ puts it into the form $\displaystyle\dfrac{1}{3}\int\dfrac{1}{t^2+1}dt$ that you should know how to manage. $\endgroup$ – egreg Oct 8 '13 at 20:09
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You know (or should know) that $\int\frac{1}{x^2+1}\mathrm{dx}=\arctan(x)$. Let's try and get the integrand into that form.

$$\int\frac{1}{x^2+9}\mathrm{dx}=\int\frac{1}{9(\frac{x^2}{9}+1)}\mathrm{dx}=\frac{1}{9}\int\frac{1}{\left(\frac{x}{3}\right)^2+1}\mathrm{dx}$$

You also know (or should know) that you can easily substitute out that $x/3$:

$$=\frac{1}{9}3\int\frac{1}{\left(\frac{x}{3}\right)^2+1}\frac{\mathrm{dx}}{3}$$

$$=\frac{1}{3}\arctan(\frac{x}{3})$$

Any time you've got an integrand of the form 1 divided by some quadtratic, you can usually (not always) get it into that form with a bit of fiddling, sometimes involving completing the square.

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Make the substitution

$$ x=3\tan t \implies dx = 3\sec^2 t \,dt .$$

Subs back in the integral and you need to use the identity

$$ 1+\tan^2 t = \sec^2 t. $$

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Check this table of integrals

Integral number 9 is the one you need, where the value of a = 3

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