Problem 70 - Calculus for Biology and Medicine

Question

Evaluate each definite integral. $$ \int_{0}^{\pi / 6}\left(1+\tan ^{2} x\right) d x $$

Step-by-Step Solution

Answer

\(\frac{1}{\sqrt{3}}\)

Step 1: Identify the Integral

The given integral is \( \int_{0}^{\pi / 6}(1+\tan^{2}x)\,dx \). The integrand is \(1 + \tan^2 x\), which we recognize from the trigonometric identity as \(\sec^2 x\), since \(1 + \tan^2 x = \sec^2 x\).

Step 2: Simplify using Trigonometric Identity

Using the identity \(1 + \tan^2 x = \sec^2 x\), rewrite the integral as \( \int_{0}^{\pi / 6} \sec^2 x\,dx \).

Step 3: Integrate the Simplified Function

The antiderivative of \(\sec^2 x\) is \(\tan x\). So, the integral \( \int \sec^2 x\,dx \) simplifies to \(\tan x + C\), where \(C\) is the constant of integration.

Step 4: Evaluate the Indefinite Integral over Limits

Evaluate the antiderivative from \(0\) to \(\pi/6\). This gives us \(\tan\left(\frac{\pi}{6}\right) - \tan(0)\).

Step 5: Compute the Trigonometric Values

Calculate \(\tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}\) and \(\tan(0) = 0\).

Step 6: Final Calculation

Substitute the values into the integral to find \(\frac{1}{\sqrt{3}} - 0 = \frac{1}{\sqrt{3}}\). Therefore, the value of the definite integral is \(\frac{1}{\sqrt{3}}\).

Key Concepts

Definite Integrals

A definite integral is a fundamental concept in calculus, representing the area under a curve within specific limits on a graph. For the given problem, our definite integral is \( \int_{0}^{\pi / 6}(1+\tan^2 x)\,dx \). This means we're looking for the total area under the function \(1 + \tan^2 x\) from \(x = 0\) to \(x = \frac{\pi}{6}\).

What makes definite integrals so powerful? They allow us to calculate the total accumulation of quantities that vary over time, such as area, volume, or even probability. The notation \(\int_a^b f(x) \,dx\) tells us to find the antiderivative of \(f(x)\), which becomes crucial when doing the calculations.

Definite integrals have two main features:

• Upper and lower limits (in our exercise, \(0\) and \(\frac{\pi}{6}\)).
• They provide a specific result, not a general formula with a constant \(C\).

This leads us to calculate the net or total change in the quantity described by the function over the interval \([a, b]\).

Trigonometric Identities

Understanding trigonometric identities is key to simplifying integrals involving trigonometric functions. In our exercise, the integral involves \(1 + \tan^2 x\).

Recognizing the identity
\[ 1 + \tan^2 x = \sec^2 x \]
is crucial. It simplifies calculations by transforming the function into something more straightforward to work with. By applying this identity, our integral becomes \( \int_{0}^{\pi / 6} \sec^2 x \,dx \).

Some important trigonometric identities include:

• \(\sin^2 x + \cos^2 x = 1\)
• \(1 + \tan^2 x = \sec^2 x\)
• \(1 + \cot^2 x = \csc^2 x\)

Using these identities simplifies integration and differentiation, making our calculations more efficient and manageable.

Antiderivatives

Antiderivatives, or indefinite integrals, are functions that represent the "reverse" process of differentiation. For any function \(f(x)\), the antiderivative is another function \(F(x)\) such that the derivative of \(F(x)\) is \(f(x)\).

In our given problem, we need to find the antiderivative of \(\sec^2 x\) to compute the definite integral. Recognizing that the derivative of \(\tan x\) is \(\sec^2 x\), we know:
\[ \int \sec^2 x \,dx = \tan x + C \]

Where \(C\) is the constant of integration, which we exclude when calculating a definite integral.

Once you find the antiderivative, you substitute the upper and lower limits of the definite integral into the antiderivative function to find the precise solution. That's why for our specific problem, we evaluated \(\tan\left(\frac{\pi}{6}\right) - \tan(0)\) to compute the result.