Problem 125 - Calculus Volume 1

Question

Suppose that \(A=\int_{-\pi / 4}^{\pi / 4} \sec ^{2} t d t=\pi\) and \(B=\int_{-\pi / 4}^{\pi / 4} \tan ^{2} t d t\). Show that \(A-B=\frac{\pi}{2}\)

Step-by-Step Solution

Answer

Thus, \(A - B = \frac{\pi}{2}\).

Step 1: Understand the given integrals

We are given two integrals \(A = \int_{-\pi/4}^{\pi/4} \sec^2 t \, dt\) and \(B = \int_{-\pi/4}^{\pi/4} \tan^2 t \, dt\). We know that \(A = \pi\). Our goal is to find the expression for \(A - B\) and show that it equals \(\frac{\pi}{2}\).

Step 2: Use trigonometric identity

Recall the trigonometric identity \(\sec^2 t = 1 + \tan^2 t\). This identity can be used to express \(\sec^2 t\) in terms of \(\tan^2 t\), allowing us to relate integrals \(A\) and \(B\).

Step 3: Substitute the identity into integral A

Using the identity \(\sec^2 t = 1 + \tan^2 t\), substitute into integral \(A\) to get:\[A = \int_{-\pi/4}^{\pi/4} (1 + \tan^2 t) \, dt = \int_{-\pi/4}^{\pi/4} 1 \, dt + \int_{-\pi/4}^{\pi/4} \tan^2 t \, dt\]

Step 4: Separate the integral

Separate the integral into two parts:\[A = \int_{-\pi/4}^{\pi/4} 1 \, dt + \int_{-\pi/4}^{\pi/4} \tan^2 t \, dt\]The first integral evaluates to the length of the interval \([-\pi/4, \pi/4]\), which is \(\pi/2\).

Step 5: Express A in terms of B

From the separation, we have:\[A = \frac{\pi}{2} + B\]We know from the given information that \(A = \pi\).

Step 6: Solve for A-B

Using \(A = \frac{\pi}{2} + B\), substitute \(A = \pi\) into the expression to find:\[\pi = \frac{\pi}{2} + B\]Solving for \(B\) gives:\[B = \pi - \frac{\pi}{2} = \frac{\pi}{2}\]Thus, \(A - B = \pi - \frac{\pi}{2} = \frac{\pi}{2}\).

Step 7: Verify the result

Verify that combining our expressions correctly simplifies to show that \(A - B = \frac{\pi}{2}\), matching the requirement given in the problem statement.

Key Concepts

Trigonometric Substitution

Trigonometric substitution is a powerful technique used in calculus to simplify integrals. When dealing with integrals that involve square roots or certain trigonometric functions, substituting using a trigonometric identity can make the equation easier to solve. In the given problem, we worked with the integral of \(\sec^2 t\) and \(\tan^2 t\), which are common in many trigonometric substitution problems.
In our exercise, the substitution isn't typical like in problems involving \(\sqrt{a^2 - x^2}\) where you'd use \(x = a \sin\theta\). Instead, it's about utilizing the identity \(\sec^2 t = 1 + \tan^2 t\) directly to simplify the integral. This identity allowed us to break down the complex integral into simpler parts, showing how one can relate different functions using trigonometric principles. By substituting and recognizing known relationships, we simplify our approach, making it easier to find the solution.
Trigonometric substitutions are especially useful when you can see the direct application of identities. Therefore, understanding and recognizing these identities and how they relate can significantly help in solving complex calculus problems.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are true for every value of the occurring variables where both sides of the equality are defined. These are critical in calculus when dealing with integrals of trigonometric functions. In the problem we've been solving, one such identity, \(\sec^2 t = 1 + \tan^2 t\), proved instrumental.
The identity \(\sec^2 t = 1 + \tan^2 t\) comes from the Pythagorean identity \(\sec^2 t - \tan^2 t = 1\). This specific identity allowed us to express \(\sec^2 t\) in terms of \(\tan^2 t\). Using this identity, we were able to break down the integral \(A = \int_{-\pi/4}^{\pi/4} \sec^2 t \, dt\) into the sum of two simpler integrals: \(\int_{-\pi/4}^{\pi/4} 1 \, dt\) and \(\int_{-\pi/4}^{\pi/4} \tan^2 t \, dt\). This manipulation was key to solving the problem, showcasing the power of trigonometric identities in simplifying and solving calculus problems involving trigonometric expressions.
Understanding and applying these identities can simplify many calculus tasks, making them invaluable to students learning integration.

Integration Techniques

Integration techniques are strategies used to solve integrals, which are essential in calculus for finding areas, solving differential equations, and many other applications. One common integration technique involves using trigonometric identities and substitutions, as in this exercise.
Our problem demonstrates a straightforward integration technique: evaluating the definite integral of a constant function, \(\int 1 \, dt\), which equaled the length of the interval \(\pi/2\). This particular evaluation simplified the given integral \(A\) in terms of \(B\). This simplification step highlighted direct integration as an effective technique in conjunction with the application of identities. The problem also showed how we can methodically subtract these results to find differences between integrals, helping show that \(A-B=\frac{\pi}{2}\).
To master integration:

• Understand basic integrals, e.g., \(\int 1 \, dt\)
• Employ trigonometric identities for simplification
• Make use of substitutions when necessary

By learning these techniques, one can tackle a wide range of integrals in calculus, making them crucial tools for students.