Solving ∫ Sec(t)(9sec(t) + 2tan(t)) Dt A Step-by-Step Guide

Hey everyone! Today, we're diving into an exciting integral problem. We've got a trigonometric function inside an integral, and our mission, should we choose to accept it, is to find the general indefinite integral. Remember, that sneaky "constant of integration" will be making an appearance, so keep an eye out! Let's get started and make some math magic happen!

Understanding the Integral

The integral we need to solve is:

$$\int \sec (t)(9 \sec (t)+2 \tan (t)) d t$$

At first glance, this might seem a bit intimidating. But don't worry, guys! We're going to break it down step by step. The key here is to recognize that we can simplify this by distributing the sec(t) term. This will give us a sum of terms that we can integrate individually. Think of it like untangling a knot – we're taking something complex and making it manageable. Remember, in calculus, often the most challenging-looking problems are just a few clever steps away from a straightforward solution. So, keep your chin up, and let's tackle this bad boy together!

Breaking Down the Problem

So, our first step is to distribute the sec(t) across the terms inside the parentheses. This means we'll be multiplying sec(t) by both 9sec(t) and 2tan(t). When we do this, we get:

$$\int (9 \sec^2 (t) + 2 \sec (t) \tan (t)) dt$$

Now, this looks much better, doesn't it? We've transformed our original integral into the sum of two simpler integrals. This is a classic technique in calculus – break a complex problem into smaller, more manageable pieces. It's like conquering a mountain: you don't try to climb it in one giant leap, you take it one step at a time. And that's exactly what we're doing here. We're turning one tough integral into two easier ones. Remember, the power of calculus often lies in its ability to simplify and break down problems. So, let's keep this momentum going and see how we can integrate these individual terms.

Integrating the Terms

Now we have two separate integrals to deal with:

$$\int 9 \sec^2 (t) dt + \int 2 \sec (t) \tan (t) dt$$

These integrals are actually quite common, and you might even recognize them from your calculus studies. The first one involves sec²(t), and the second one involves the product of sec(t) and tan(t). If you've memorized your basic integral formulas (and if you haven't, now's a great time to brush up!), you'll know that these have direct antiderivatives.

Let's focus on the first integral: ∫9sec²(t) dt. Remember that the derivative of tan(t) is sec²(t). This means that the integral of sec²(t) is simply tan(t). And since we have a constant multiple of 9, we just carry that along. So, the integral of 9sec²(t) is 9tan(t).

Now, let's tackle the second integral: ∫2sec(t)tan(t) dt. Recall that the derivative of sec(t) is sec(t)tan(t). This means the integral of sec(t)tan(t) is sec(t). Again, we have a constant multiple, this time 2, so we carry that along as well. Thus, the integral of 2sec(t)tan(t) is 2sec(t).

See? We're making great progress! By recognizing these standard integral forms, we've been able to directly find the antiderivatives of our terms. This highlights the importance of knowing your basic calculus formulas – they can save you a lot of time and effort. Now that we've integrated each term individually, we're just one step away from the final answer.

The Constant of Integration

We've found the antiderivatives of both parts of our integral, which gives us:

$$9 \tan (t) + 2 \sec (t)$$

But hold on! We're not quite finished yet. There's one crucial thing we need to remember whenever we're dealing with indefinite integrals: the constant of integration. This is a fancy way of saying that when we find an antiderivative, there could be a constant term that disappears when we take the derivative. For example, the derivative of x² + 5 is 2x, but so is the derivative of x² - 10 or even x² + π. The constant term vanishes during differentiation.

That's why, when we find an indefinite integral, we always add a constant term, usually denoted by 'C', to account for this possibility. It's like saying, "Hey, there might be a constant hiding here, so let's not forget about it!" So, our final answer isn't just 9tan(t) + 2sec(t); it's 9tan(t) + 2sec(t) + C. This 'C' is the key to representing the general indefinite integral, which includes all possible constant terms. Never forget the + C! It's the hallmark of a complete and correct indefinite integral solution.

The Final Solution

Putting it all together, the general indefinite integral is:

$$\int \sec (t)(9 \sec (t)+2 \tan (t)) d t = 9 \tan (t) + 2 \sec (t) + C$$

There you have it, guys! We've successfully navigated this integral problem. We started with what looked like a complex expression, but by distributing, breaking it into smaller integrals, recognizing standard forms, and remembering our constant of integration, we arrived at a beautiful solution. Remember, practice makes perfect, so keep tackling those integrals, and you'll become a calculus pro in no time! This problem really highlights the core techniques of integration: simplification, recognition of standard forms, and meticulous attention to detail (especially that + C!). So, next time you encounter an integral that looks a bit scary, remember this journey and break it down step by step. You've got this!

Key Takeaways

Before we wrap up, let's quickly recap the key steps we took to solve this integral. This will help solidify your understanding and give you a clear roadmap for tackling similar problems in the future. Think of these as the essential tools in your integral-solving toolbox.

1. Distribution is your friend: The first step was to distribute the sec(t) term. This is a common strategy in integration – simplifying the expression inside the integral can make it much easier to work with. Look for opportunities to multiply, expand, or otherwise simplify the integrand before diving into integration.

2. Break it down: We broke the integral into a sum of simpler integrals. This is a powerful technique for dealing with complex integrands. If you can express your integral as a sum or difference of terms, you can integrate each term separately. This often transforms a daunting problem into a series of more manageable ones.

3. Know your standard forms: Recognizing common integral forms is crucial. We relied on our knowledge of the integrals of sec²(t) and sec(t)tan(t). Building a strong foundation of basic integral formulas is essential for success in calculus. Flashcards, practice problems, and regular review can help you master these key forms.

4. Don't forget + C: Always remember the constant of integration! It's a small detail, but it's incredibly important for representing the general indefinite integral. Make it a habit to add '+ C' to the end of every indefinite integral you solve. Think of it as the final flourish on your masterpiece of calculus.

By keeping these takeaways in mind, you'll be well-equipped to tackle a wide range of integral problems. Remember, calculus is a skill that improves with practice, so keep working at it, and you'll see your abilities grow. And who knows, maybe you'll even start to enjoy the challenge of finding those integrals!

I hope this explanation has been helpful, guys! Keep practicing, and you'll be conquering integrals like a pro in no time. Until next time, happy integrating!