Problem 37 - Calculus: One and Several Variables

Question

Calculate. $$\int_{0}^{\pi / 4} \frac{1+\sin x}{\cos ^{2} x} d x$$

Step-by-Step Solution

Answer

The value of the given integral is \(2 - \sqrt{2}\).

Step 1: 1. Integrate the first part

To integrate \(\frac{1}{\cos^2 x}\), we recognize that it is the derivative of the tangent function: \(\frac{d}{dx} \tan(x) = \sec^2(x) = \frac{1}{\cos^2 x}\) So, the integral of the first part is: \(\int_{0}^{\pi/4} \frac{1}{\cos^2 x}\, dx = \int_{0}^{\pi/4} \sec^2(x)\, dx = \tan(x) \Big|_0^{\pi/4}\) Evaluating the tangent function at the integration limits, we get \(\tan \left(\frac{\pi}{4}\right) - \tan(0) = 1 - 0 = 1\) So, the integral of the first part is equal to 1.

Step 2: 2. Integrate the second part

To integrate \(\frac{\sin x}{\cos^2 x}\), we use substitution. Let's use a substitution: \(u = \cos(x) \Rightarrow du = -\sin(x)dx\) Now, we rewrite the integral as follows: \(\int_{0}^{\pi/4} \frac{\sin x}{\cos^2 x}\, dx =\int_{1}^{\frac{1}{\sqrt{2}}} -\frac{1}{u^2} du\) Now, we can integrate: \(-\int_{1}^{\frac{1}{\sqrt{2}}} \frac{1}{u^2} du = -\left[ \frac{-1}{u} \right]_1^{\frac{1}{\sqrt{2}}}\) Evaluating the integral at the integration limits, we get: \(-\left( \frac{-1}{\frac{1}{\sqrt{2}}} - \frac{-1}{1}\right) = -\left( \sqrt{2} - 1\right) = 1 - \sqrt{2}\) So, the integral of the second part is equal to \(1 - \sqrt{2}\).

Step 3: 3. Combine the results

Now, we combine the results of the two integrals: \(\int_{0}^{\pi/4} \frac{1+\sin x}{\cos^2 x}\, dx = 1 + (1 - \sqrt{2}) = 2 - \sqrt{2}\) Therefore, the value of the given integral is \(2 - \sqrt{2}\).

Key Concepts

Understanding Trigonometric Integrals

Trigonometric integrals involve the integration of functions that contain trigonometric expressions such as sine, cosine, or tangent. These integrals often require special techniques like substitution to simplify and solve.
In our example, the integral \( \int_{0}^{\pi / 4} \frac{1+\sin x}{\cos ^{2} x} \, dx \) requires breaking down into parts with familiar trigonometric derivatives.
Recognizing the derivatives of trigonometric functions helps simplify complex expressions, making them easier to integrate.

Exploring Definite Integrals

A definite integral calculates the net area between a function and the x-axis, within specific limits. Unlike indefinite integrals that include a constant of integration, definite integrals provide a particular value.
In our integral, the limits from \(0\) to \(\pi/4\) mean we find the area under the curve between these points.

• The Fundamental Theorem of Calculus tells us to evaluate the antiderivative at these upper and lower limits, then subtract the results.
• This process ensures an exact result, as seen in the integration of \(\frac{1}{\cos^2 x}\) from \(0\) to \(\pi/4\).

Simplifying with the Substitution Method

The substitution method is a powerful technique to simplify integrals by changing variables. It converts complex integrals into easier forms.
In our example, using \(u = \cos(x)\) changes the integral \(\int \frac{\sin x}{\cos^2 x} \, dx\) into a more manageable form.

• This method requires substituting \(du\) for \(-\sin(x)dx\) and adjusting limits accordingly.
• By rewriting the integral in terms of \(u\), we make the integration process straightforward and intuitive.

Working with the Tangent Function

The tangent function, \(\tan(x)\), plays a key role in trigonometric integrals. Its derivative, \(\sec^2(x)\), is particularly useful.
In our problem, recognizing that \(\frac{1}{\cos^2 x}\) equals \(\sec^2(x)\) allows direct integration to \(\tan(x)\).

• This insight simplifies the initial part of the integral quickly, making grinding through multiple steps unnecessary.
• The function \(\tan(x)\) at the limits provides direct calculation, demonstrating why understanding fundamental trigonometric identities is beneficial.