# Need help evaluating the definite integral.

$$\int_0^6 \frac{dx}{x^2+36}$$

This question is killing me. Ive done it 5 different times and not one answer was right.

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The denominator signals a tangent substitution, $x=6\tan\theta$; that’s one of the basic, standard trig substitutions. Then $dx=6\sec^2\theta$, $\tan 0=0$, and $6\tan\frac{\pi}4=6$, so your integral becomes

$$\int_0^{\pi/4}\frac{6\sec^2\theta~d\theta}{36\tan^2\theta+36}=\frac16\int_0^{\pi/4}\frac{\sec^2\theta~d\theta}{\tan^2\theta+1}=\frac16\int_0^{\pi/4}d\theta\;,$$

which is a very easy integral.

Alternatively, notice that

$$\frac1{x^2+36}=\frac1{36\left(\left(\frac{x}6\right)^2+1\right)}\;,$$

so your integral can be rewritten as

$$\frac1{36}\int_0^6\frac{dx}{1+(x/6)^2}\;.\tag{1}$$

You should recognize that as basically the derivative of an arctangent. Let $u=\frac{x}6$, so that $du=\frac16$, and you can rewrite $(1)$ as

$$\frac16\int_0^1\frac{du}{1+u^2}=\frac16\left[\tan^{-1}x\right]_0^1\;,$$

which is also easy to evaluate.

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my answer came out to pi/24. Is that correct? – Ak47 Feb 7 '13 at 4:38
@Ak47: Yes, it is. – Brian M. Scott Feb 7 '13 at 4:43
thank you so much. – Ak47 Feb 7 '13 at 4:50
@Ak47: You’re very welcome. – Brian M. Scott Feb 7 '13 at 4:53

Let $x=6 \tan{t}$, $dx=6 \sec^2{t} \, dt$, and note that $1+\tan^2{t} = \sec^2{t}$.

The integral is then

\begin{align}\int_0^6 \frac{dx}{x^2+36}&= \frac{6}{36} \int_0^{\pi/4} dt \frac{\sec^2{t}}{\sec^2{t}} \\ &= \frac{\pi}{24}\end{align}

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Why would x = 6tan t? – Ak47 Feb 7 '13 at 4:13
This is what is known as a trig substitution, where we look for patterns in the integrand that resemble trig identities. Patterns such as $x^2 + a^2$ in the integrand frequently are addressed by using a substitution $x = \tan{t}$. – Ron Gordon Feb 7 '13 at 4:17
Also note the effect of the substitution in the integral limits. $t=\arctan{x/6}$, so $\arctan{6/6} = \arctan{1} = \pi/4$. – Ron Gordon Feb 7 '13 at 4:20

It fits the form $a^2 + u^2$ where $a \in \mathbb R$. Thus you should use $x = 6 \tan \theta \implies dx = 6\sec^2 \theta \ d\theta.$ You should be able to finish it off now by recalling $1+\tan^2 x = \sec^2 x.$

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In general, for any derivable function $\,f(x)\,$ :

$$\int\frac{f'(x)dx}{1+f(x)^2}=\arctan f(x)+C$$

This makes your integral almost immediate, and without any need to make substitutions:

$$\int\frac{dx}{36+x^2}=\frac{1}{6}\int\frac{\frac{1}{6}dx}{1+\left(\frac{x}{6}\right)^2}=\frac{1}{6}\arctan\frac{x}{6}+C$$

since $\,\frac{1}{6}=\left(\frac{x}{6}\right)'\,$

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+1 I love this small latter identity, Don. I feel that, it has shaken the World many times, many...But we always neglect it and its rule. – Babak S. Feb 8 '13 at 19:06

Here's different way from what others have posted so far. $$\int_0^6 \frac{dx}{x^2+36} = \frac16\int_0^6 \frac{dx/6}{\left(\frac x6\right)^2 + 1} = \frac16\int_0^1 \frac{du}{x^2+1} = \frac16(\arctan1-\arctan0) = \frac16\cdot\frac\pi4.$$

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