Problem 25 - Biocalculus Calculus for the Life Sciences

Question

Evaluate the integral. \(\int_{0}^{\pi / 4} \frac{1+\cos ^{2} \theta}{\cos ^{2} \theta} d \theta\)

Step-by-Step Solution

Answer

The integral evaluates to \(1 + \frac{\pi}{4}\).

Step 1: Simplify the Integrand

First, simplify the integrand \(\frac{1 + \cos^2 \theta}{\cos^2 \theta}\). This can be rewritten as \(\frac{1}{\cos^2 \theta} + 1\) by splitting the fraction. The expression \(\frac{1}{\cos^2 \theta}\) is the same as \(\sec^2 \theta\). Therefore, the integrand becomes \(\sec^2 \theta + 1\).

Step 2: Set Up the Integral

With the simplified integrand, the integral is now \(\int_{0}^{\pi / 4} (\sec^2 \theta + 1) \, d\theta\). This involves the integral of \(\sec^2 \theta\) and the integral of \(1\) over the interval \([0, \pi / 4]\).

Step 3: Integrate \(\sec^2 \theta\)

The integral of \(\sec^2 \theta\) with respect to \(\theta\) is \(\tan \theta\). Therefore, \(\int \sec^2 \theta \, d\theta = \tan \theta + C\).

Step 4: Integrate Constant Term

The integral of \(1\) with respect to \(\theta\) is simply \(\theta\). So, \(\int 1 \, d\theta = \theta + C\).

Step 5: Evaluate the Definite Integral

Substitute the antiderivatives into the definite integral: \(\left[ \tan \theta + \theta \right]_{0}^{\pi / 4}\). Calculate \(\tan(\pi/4) + \pi/4 - (\tan(0) + 0)\).\[\tan(\pi/4) = 1\] \[ \tan(0) = 0\].

Step 6: Final Calculation

Evaluate the expression: \(1 + \pi/4 - 0 - 0 = 1 + \pi/4\).

Step 7: Conclude the Solution

Thus, the value of the integral \(\int_{0}^{\pi/4} \frac{1+\cos^2 \theta}{\cos^2 \theta} \, d\theta \) is \(1 + \frac{\pi}{4}\).

Key Concepts

Integration

Integration is a fundamental concept in calculus used to find areas under curves, volumes, and other quantities that accumulate. In our specific problem, we're working with definite integrals, which provide the net area under a curve described by the function between given limits. To solve integral problems like the one in the exercise, it's crucial to simplify the integrand first. This involves rewriting and breaking down the function, so it's easier to manage.
In our instance, the expression was simplified from \( \frac{1 + \cos^2 \theta}{\cos^2 \theta} \) to \( \sec^2 \theta + 1 \). This simplification is crucial as it allows us to integrate standard functions easily. Integration involves finding an antiderivative, a reverse operation of differentiation. Each function has its own integration technique, and familiarity with common ones is vital for solving these problems efficiently.
Common integration techniques include:

• Substitution: Changing variables to simplify the integrand.
• Integration by parts: Useful for products of functions.
• Partial fraction decomposition: Breaks down complex rational expressions.

Understanding and practicing these techniques helps tackle a wide array of integral problems.

Trigonometric Functions

Trigonometric functions like sine, cosine, and tangent are essential in calculus and the study of integration. For instance, in the exercise, we simplified the integrand to \( \sec^2 \theta \), which is related to the trigonometric function secant. The secant is the reciprocal of the cosine function, \( \sec\theta = \frac{1}{\cos\theta} \). Trigonometric identities often help in simplifying and solving integrals.
In this problem, we encounter the function \( \sec^2 \theta \), which is the derivative of the tangent function. This makes finding the integral straightforward. The key trigonometric functions and their derivatives you should be familiar with are:

• \( \frac{d}{d\theta} (\sin\theta) = \cos\theta \)
• \( \frac{d}{d\theta} (\cos\theta) = -\sin\theta \)
• \( \frac{d}{d\theta} (\tan\theta) = \sec^2\theta \)
• \( \frac{d}{d\theta} (\sec\theta) = \sec\theta \tan\theta \)

Remembering these derivatives allows you to find antiderivatives, or reverse the processes, when integrating.

Definite Integral

Definite integrals calculate the net area under a curve between two points, providing a precise numerical result rather than a formula. In our exercise, we integrated \( \sec^2 \theta + 1 \) over the interval \([0, \pi/4]\). After finding the antiderivative, which was \( \tan \theta + \theta \), it was time to evaluate it at the bounds.
To do this, we applied the fundamental theorem of calculus, which connects differentiation and integration, stating that if \( F(x) \) is an antiderivative of \( f(x) \), then:\[ \int_{a}^{b} f(x) \, dx = F(b) - F(a). \]We evaluated \( \tan \theta + \theta \) at \( \pi/4 \) and \( 0 \), and subtracted the two results, finding the net area under the curve between these angles. In our problem:

• At \( \theta = \pi/4 \), \( \tan(\pi/4) = 1 \) and the result is \( 1 + \pi/4 \).
• At \( \theta = 0 \), \( \tan(0) = 0 \) and the result is \( 0 \).

The difference provides the final answer: \( 1 + \frac{\pi}{4} \). Understanding the definite integral involves grasping this connection between antiderivatives and the limits of integration.