Solve Definite Integrals: A Step-by-Step Guide

Introduction

Hey guys! Today, we're diving deep into the fascinating world of definite integrals, specifically tackling a problem that involves some trigonometric functions and clever substitutions. We'll break down the problem step-by-step, making sure everyone, from calculus newbies to seasoned math enthusiasts, can follow along. Our main focus will be on evaluating the integrals $u=\displaystyle\int_0^{\pi/2}\cos\left(\frac{2\pi}3\sin^2(x)\right)\,\mathrm dx$ and $v=\displaystyle\int_0^{\pi/2}\cos\left(\frac\pi3\sin(x)\right)\,\mathrm dx$ and finding the relationship between them. So, buckle up and let's get started!

Definite integrals are a cornerstone of calculus, representing the signed area under a curve between two specified limits. They pop up in various fields, from physics and engineering to economics and computer science. Mastering the art of solving definite integrals is crucial for anyone looking to build a strong foundation in these areas. The problem we're tackling today beautifully illustrates how trigonometric identities, substitution techniques, and a bit of mathematical intuition can come together to solve seemingly complex integrals. We'll explore the underlying concepts, the step-by-step solution, and the elegant relationship that emerges between u and v. By the end of this guide, you'll not only understand how to solve this specific problem but also gain valuable insights into tackling similar challenges in the realm of definite integrals. We will use a blend of explanation, calculations, and practical tips to ensure a comprehensive understanding. Remember, the key to mastering calculus is practice, so feel free to try out similar problems and explore different techniques. Let’s embark on this mathematical journey together and unlock the secrets hidden within these integrals.

Problem Statement

Before we jump into the solution, let's clearly state the problem we're going to solve. We are given two definite integrals:

$u=\displaystyle\int_0^{\pi/2}\cos\left(\frac{2\pi}3\sin^2(x)\right)\,\mathrm dx$

and

$v=\displaystyle\int_0^{\pi/2}\cos\left(\frac\pi3\sin(x)\right)\,\mathrm dx$

Our mission, should we choose to accept it (and we do!), is to determine the relationship between u and v. This means we need to find an equation or a connection that links these two integrals. This might involve showing that one is a multiple of the other, that they are equal, or some other interesting relationship. To crack this, we'll need to roll up our sleeves and employ a combination of trigonometric identities, integral properties, and perhaps even a clever substitution or two. The beauty of this problem lies in the way it brings together different calculus concepts to arrive at a concise and elegant solution. We are not just looking for numerical values; we aim to uncover the intrinsic link between these two integrals, which is a more profound and rewarding mathematical endeavor. So, let’s keep the problem firmly in mind as we proceed, and watch how each step brings us closer to unveiling the relationship between u and v. Remember, the journey of solving a mathematical problem is often as enlightening as the solution itself.

Solution Approach

Okay, let’s map out our strategy for solving this definite integral problem. The first thing we should consider when faced with integrals involving trigonometric functions is whether we can use any trigonometric identities to simplify the integrand. In the case of u, we have $\cos\left(\frac{2\pi}3\sin^2(x)\right)$, which might benefit from a double-angle formula or a Pythagorean identity. For v, the term $\cos\left(\frac\pi3\sin(x)\right)$ looks like it might be trickier to directly simplify with basic identities. The appearance of $\sin^2(x)$ in the integral u suggests that we could potentially use the identity $\sin^2(x) = \frac{1 - \cos(2x)}{2}$ to rewrite the integrand. This substitution might make the integral more manageable. Another avenue to explore is special integral forms. Sometimes, definite integrals can be linked to known integral results, such as those involving Bessel functions or other special functions. If we can massage our integrals into a familiar form, we might be able to directly evaluate them or relate them to each other. Given the complexity of the integrands, we might also consider using series expansions. Recall that trigonometric functions like cosine can be expressed as infinite series. Substituting these series into the integrals might allow us to evaluate the integrals term by term, or at least identify a pattern that reveals the relationship between u and v. Finally, we'll need to carefully manipulate the resulting expressions and use our algebraic skills to uncover the connection between u and v. Remember, sometimes the solution isn't immediately obvious, and it requires a bit of mathematical experimentation and intuition. So, let’s dive in and see which of these approaches, or perhaps a combination of them, will lead us to the answer!

Step-by-Step Solution

Let's begin by tackling the integral u. As we discussed, the presence of $\sin^2(x)$ inside the cosine function hints at using the identity $\sin^2(x) = \frac{1 - \cos(2x)}{2}$. Let's substitute this into the expression for u:

$u = \displaystyle\int_0^{\pi/2}\cos\left(\frac{2\pi}3\sin^2(x)\right)\,\mathrm dx = \displaystyle\int_0^{\pi/2}\cos\left(\frac{2\pi}3\cdot\frac{1 - \cos(2x)}{2}\right)\,\mathrm dx$

Simplifying the argument of the cosine, we get:

$u = \displaystyle\int_0^{\pi/2}\cos\left(\frac{\pi}3(1 - \cos(2x))\right)\,\mathrm dx$

Now, we can expand the argument:

$u = \displaystyle\int_0^{\pi/2}\cos\left(\frac{\pi}3 - \frac{\pi}3\cos(2x)\right)\,\mathrm dx$

At this point, we can use the cosine subtraction formula, $\cos(A - B) = \cos(A)\cos(B) + \sin(A)\sin(B)$, where $A = \frac{\pi}3$ and $B = \frac{\pi}3\cos(2x)$:

$u = \displaystyle\int_0^{\pi/2}\left[\cos\left(\frac{\pi}3\right)\cos\left(\frac{\pi}3\cos(2x)\right) + \sin\left(\frac{\pi}3\right)\sin\left(\frac{\pi}3\cos(2x)\right)\right]\,\mathrm dx$

We know that $\cos(\frac{\pi}3) = \frac12$ and $\sin(\frac{\pi}3) = \frac{\sqrt{3}}2$. Substituting these values, we have:

$u = \displaystyle\int_0^{\pi/2}\left[\frac12\cos\left(\frac{\pi}3\cos(2x)\right) + \frac{\sqrt{3}}2\sin\left(\frac{\pi}3\cos(2x)\right)\right]\,\mathrm dx$

Now, let's make a substitution to simplify things further. Let $t = 2x$, so $dt = 2 dx$ and $dx = \frac12 dt$. When $x = 0$, $t = 0$, and when $x = \frac{\pi}2$, $t = \pi$. Thus, our integral becomes:

$u = \frac12\displaystyle\int_0^{\pi}\left[\frac12\cos\left(\frac{\pi}3\cos(t)\right) + \frac{\sqrt{3}}2\sin\left(\frac{\pi}3\cos(t)\right)\right]\,\mathrm dt$

Now, let's consider the integral v. We have:

$v = \displaystyle\int_0^{\pi/2}\cos\left(\frac\pi3\sin(x)\right)\,\mathrm dx$

To connect u and v, we can use the property of definite integrals that $\displaystyle\int_0^{2a}f(x)\,\mathrm dx = 2\displaystyle\int_0^af(x)\,\mathrm dx$ if $f(2a - x) = f(x)$. Let's see if we can apply this to our integral. In our case, we need to check if:

$\cos\left(\frac{\pi}3\cos(\pi - t)\right) = \cos\left(\frac{\pi}3\cos(t)\right)$

Since $\cos(\pi - t) = -\cos(t)$, the integral becomes:

$\cos\left(-\frac{\pi}3\cos(t)\right) = \cos\left(\frac{\pi}3\cos(t)\right)$

This is indeed true because $\cos(-x) = \cos(x)$. So, we can split the integral u into two equal parts:

$\displaystyle\int_0^{\pi}\cos\left(\frac{\pi}3\cos(t)\right)\,\mathrm dt = 2\displaystyle\int_0^{\pi/2}\cos\left(\frac{\pi}3\cos(t)\right)\,\mathrm dt$

Now, let's substitute $x = \frac{\pi}2 - x$ in the integral v:

$v = \displaystyle\int_0^{\pi/2}\cos\left(\frac\pi3\sin(x)\right)\,\mathrm dx = \displaystyle\int_0^{\pi/2}\cos\left(\frac\pi3\sin(\frac{\pi}2 - x)\right)\,\mathrm dx = \displaystyle\int_0^{\pi/2}\cos\left(\frac\pi3\cos(x)\right)\,\mathrm dx$

Comparing this with the integral we obtained from u, we can see a clear relationship emerging.

Unveiling the Relationship

From our step-by-step solution, we've arrived at a crucial point where we can finally unveil the relationship between u and v. Recall that we transformed u into:

$u = \frac12\displaystyle\int_0^{\pi}\left[\frac12\cos\left(\frac{\pi}3\cos(t)\right) + \frac{\sqrt{3}}2\sin\left(\frac{\pi}3\cos(t)\right)\right]\,\mathrm dt$

And we used the property of even functions to simplify the integral involving the cosine term. However, let's take a closer look at the sine term: $\sin\left(\frac{\pi}3\cos(t)\right)$. The sine function is odd, meaning $\sin(-x) = -\sin(x)$. Since $\cos(\pi - t) = -\cos(t)$, we have:

$\sin\left(\frac{\pi}3\cos(\pi - t)\right) = \sin\left(-\frac{\pi}3\cos(t)\right) = -\sin\left(\frac{\pi}3\cos(t)\right)$

This means that the integral of the sine term over the interval $[0, \pi]$ is zero:

$\displaystyle\int_0^{\pi}\sin\left(\frac{\pi}3\cos(t)\right)\,\mathrm dt = 0$

Therefore, the expression for u simplifies to:

$u = \frac12\displaystyle\int_0^{\pi}\frac12\cos\left(\frac{\pi}3\cos(t)\right)\,\mathrm dt = \frac14\displaystyle\int_0^{\pi}\cos\left(\frac{\pi}3\cos(t)\right)\,\mathrm dt$

Now, using the property of even functions again, we have:

$u = \frac14 \cdot 2 \displaystyle\int_0^{\pi/2}\cos\left(\frac{\pi}3\cos(t)\right)\,\mathrm dt = \frac12\displaystyle\int_0^{\pi/2}\cos\left(\frac{\pi}3\cos(x)\right)\,\mathrm dx$

(We've changed the variable from t to x in the last step since it's a definite integral).

Comparing this to our expression for v:

$v = \displaystyle\int_0^{\pi/2}\cos\left(\frac\pi3\sin(x)\right)\,\mathrm dx = \displaystyle\int_0^{\pi/2}\cos\left(\frac\pi3\cos(x)\right)\,\mathrm dx$

We can see a clear relationship: u = (1/2)v. Isn't that neat? All that trigonometric manipulation and integration, and we find that u is simply half of v! This highlights the power of using trigonometric identities and integral properties to simplify complex expressions and reveal underlying connections.

Conclusion

Alright guys, we've successfully navigated through this definite integral problem! We started with two seemingly complicated integrals, u and v, and through a combination of trigonometric identities, clever substitutions, and a dash of mathematical intuition, we discovered the elegant relationship between them: $u = \frac12 v$. This journey underscores the beauty and power of calculus, showing how seemingly disparate expressions can be linked through fundamental principles.

Key takeaways from this problem include:

• Trigonometric Identities are Your Friends: Don't shy away from using identities like $\sin^2(x) = \frac{1 - \cos(2x)}{2}$ and $\cos(A - B) = \cos(A)\cos(B) + \sin(A)\sin(B)$ to simplify integrands.
• Substitution is a Powerful Tool: Strategic substitutions, like $t = 2x$, can transform integrals into more manageable forms.
• Exploit Symmetry: Recognizing even and odd functions and using properties like $\displaystyle\int_0^{2a}f(x)\,\mathrm dx = 2\displaystyle\int_0^af(x)\,\mathrm dx$ (if $f(2a - x) = f(x)$) can significantly simplify calculations.
• Don't Forget the Basics: Remember fundamental integral properties and trigonometric relationships. They are the building blocks of solving more complex problems.

This problem serves as a great example of how mathematical problem-solving often involves a blend of techniques and a willingness to explore different avenues. It's not always about finding the right method immediately, but rather about trying different approaches and seeing where they lead. Remember, practice is key! The more you work through problems like this, the more comfortable you'll become with the various techniques and the more intuitive your problem-solving skills will become.

So, keep exploring, keep practicing, and keep enjoying the wonderful world of calculus! You've got this!