How To Integrate 5/(x^2+36)

What's up, math heads! Today, we're diving deep into the awesome world of calculus to tackle a definite integral that might look a little intimidating at first glance, but trust me, guys, it's totally manageable once you know the tricks. We're going to break down how to integrate $\int \frac{5}{x^2+6^2} d x$. This bad boy is a classic example that pops up in many calculus courses, and understanding it is key to unlocking more complex integration techniques. So, grab your notebooks, maybe a coffee, and let's get this integration party started!

Unpacking the Integral: The Foundation

Alright, let's get down to business with our integral: $\int \frac{5}{x^2+6^2} d x$. First off, notice that the constant '5' in the numerator can be pulled right out of the integral. This is a fundamental rule in integration – constants can be factored out. So, our integral becomes $5 \int \frac{1}{x^2+6^2} d x$. See? Already looking a bit cleaner. Now, the core of our problem lies in integrating $\frac{1}{x^2+6^2}$. This form is a dead giveaway for a specific integration technique: trigonometric substitution, or sometimes referred to as trig substitution. The reason this form screams trig substitution is because of the $x^2 + a^2$ pattern, where 'a' is a constant. In our case, $a=6$. This pattern is directly related to the Pythagorean identity in trigonometry: $\tan^2(\theta) + 1 = \sec^2(\theta)$. By making a clever substitution involving the tangent function, we can transform the denominator into something much easier to integrate.

The Power of Trigonometric Substitution

So, here's the magic move, guys: we're going to use a substitution. For integrals of the form $\int \frac{1}{x^2+a^2} d x$, the standard substitution is $x = a \tan(\theta)$. In our specific problem, $a=6$, so our substitution will be $x = 6 \tan(\theta)$. Now, this is super important: when you make a substitution, you must also find the differential $dx$ in terms of $d\theta$. To do this, we differentiate both sides of our substitution $x = 6 \tan(\theta)$ with respect to $\theta$. The derivative of $x$ with respect to $\theta$ is $\frac{dx}{d\theta}$. The derivative of $6 \tan(\theta)$ is $6 \sec^2(\theta)$. So, we have $\frac{dx}{d\theta} = 6 \sec^2(\theta)$. Rearranging this to solve for $dx$, we get $dx = 6 \sec^2(\theta) d\theta$. This $dx$ term is crucial for replacing the $dx$ in our original integral.

Transforming the Integral

Now, let's plug our substitution into the integral $5 \int \frac{1}{x^2+6^2} d x$. We have $x = 6 \tan(\theta)$ and $dx = 6 \sec^2(\theta) d\theta$. Let's substitute these into the denominator first: $x^2 + 6^2 = (6 \tan(\theta))^2 + 6^2 = 36 \tan^2(\theta) + 36$. We can factor out a 36 from this expression: $36(\tan^2(\theta) + 1)$. And here's where that Pythagorean identity comes into play, my friends! Remember that $\tan^2(\theta) + 1 = \sec^2(\theta)$. So, our denominator simplifies beautifully to $36 \sec^2(\theta)$.

Now, let's put it all together. Our integral $5 \int \frac{1}{x^2+6^2} d x$ becomes:

$5 \int \frac{1}{36 \sec^2(\theta)} (6 \sec^2(\theta) d\theta)$

Look at that! The $\sec^2(\theta)$ terms in the numerator and denominator cancel each other out. Also, the '6' in the numerator of $dx$ and the '36' in the denominator cancel down to $\frac{1}{6}$. So, the integral simplifies to:

$5 \int \frac{1}{6} d\theta$

We can pull the constant $\frac{1}{6}$ out of the integral as well, giving us:

$\frac{5}{6} \int 1 d\theta$

This is now an incredibly simple integral to solve. The integral of 1 with respect to $\theta$ is just $\theta$. So, we get:

$\frac{5}{6} \theta + C$

Where 'C' is our constant of integration. We're almost done, guys, but we're not quite finished. We need to express our answer back in terms of $x$, not $\theta$.

Back-Substitution: The Final Step

To get our answer back in terms of $x$, we need to use our original substitution: $x = 6 \tan(\theta)$. We need to solve this for $\theta$. First, divide both sides by 6: $\frac{x}{6} = \tan(\theta)$. Now, to isolate $\theta$, we take the inverse tangent (or arctangent) of both sides: $\theta = \arctan(\frac{x}{6})$.

Now, we substitute this expression for $\theta$ back into our result $\frac{5}{6} \theta + C$. This gives us our final answer:

$\frac{5}{6} \arctan(\frac{x}{6}) + C$

And there you have it! We successfully integrated $\int \frac{5}{x^2+6^2} d x$. This method of trigonometric substitution is super powerful and is the key to solving many integrals that involve expressions like $x^2+a^2$, $x^2-a^2$, or $a^2-x^2$ under the square root or in the denominator. Remember the pattern, nail the substitution, and don't forget to convert $dx$ and then convert back! Keep practicing, and you'll be integrating like a pro in no time. Calculus is all about building these foundational skills, and this integral is a fantastic stepping stone. Awesome job, everyone!

Alternative Approach: Recognizing the Standard Form

Before we wrap this up, let's talk about a shortcut, or rather, a way to recognize this integral immediately if you've done your homework! You guys know that the derivative of $\arctan(u)$ is $\frac{1}{1+u^2} \frac{du}{dx}$. A related standard integral form that you should totally memorize is:

$\int \frac{1}{x^2+a^2} d x = \frac{1}{a} \arctan(\frac{x}{a}) + C$

See how this relates to our problem? We started with $\int \frac{5}{x^2+6^2} d x$. We pulled out the constant 5, leaving us with $5 \int \frac{1}{x^2+6^2} d x$. Now, compare $\frac{1}{x^2+6^2}$ to the standard form $\frac{1}{x^2+a^2}$. It's a perfect match with $a=6$!

So, using the standard formula directly, the integral $\int \frac{1}{x^2+6^2} d x$ is equal to $\frac{1}{6} \arctan(\frac{x}{6}) + C$. Since we had that '5' factored out at the beginning, we just multiply this result by 5.

$5 \times \left( \frac{1}{6} \arctan(\frac{x}{6}) + C \right)$

This gives us:

$\frac{5}{6} \arctan(\frac{x}{6}) + 5C$

Now, a common convention in calculus is that if 'C' is an arbitrary constant, then '5C' is also just an arbitrary constant. So, we usually just rename $5C$ back to $C$ to keep things simple and elegant.

$\frac{5}{6} \arctan(\frac{x}{6}) + C$

This is the exact same result we got using the more involved trigonometric substitution method! Recognizing these standard integral forms can save you a ton of time and effort on exams and in your problem-solving. It's like having a cheat sheet built into your brain! While understanding the derivation through trig substitution is super important for building a strong calculus foundation, knowing and applying these standard forms is a mark of a seasoned calculus student. So, definitely commit these standard forms to memory, especially the arctangent one, as it appears frequently in calculus problems. It's all about building that mental toolkit, guys!

Why This Matters: Applications in Real Life

So, you might be asking yourselves,