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.
 A: 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.
A: 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 $\,\frac{1}{6}=\left(\frac{x}{6}\right)'\,$
A: 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
$$\begin{align}\int_0^6 \frac{dx}{x^2+36}&= \frac{6}{36} \int_0^{\pi/4} dt \frac{\sec^2{t}}{\sec^2{t}} \\ &= \frac{\pi}{24}\end{align}$$
A: 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 $1+\tan^2 x = \sec^2 x.$
A: Here's different way from what others have posted so far.
$$
\int_0^6 \frac{dx}{x^2+36} = \frac16\int_0^6 \frac{dx/6}{\left(\frac x6\right)^2 + 1} = \frac16\int_0^1 \frac{du}{x^2+1} = \frac16(\arctan1-\arctan0) = \frac16\cdot\frac\pi4.
$$
