Question

In Exercises \(1-20,\) find \(d y / d x\). $$y=\int_{\pi}^{\pi-x} \frac{1+\sin ^{2} u}{1+\cos ^{2} u} d u$$

Short answer

\[y' = -\frac{2 - \cos^{2}x}{1 + \cos^{2}x}\]

Step-by-step solution

Rewrite the Function

First, rewrite the integral to go from \(\pi - x\) to \(\pi\) instead of the other way around. However, remember that this requires a negative sign, because the limits of the integral are switched. This results in: \[y=-\int_{\pi-x}^{\pi} \frac{1+\sin^{2}u}{1+\cos^{2}u} du\]

Apply Fundamental Theorem of calculus Part 2

Now, apply the Fundamental theorem of calculus part 2. This theorem says that the derivative of an integral of a function from \(a(x)\) to \(b(x)\) equals the difference between the function evaluated at the upper and lower limits, each multiplied by the derivative of that limit. Here, \(a(x) = \pi - x\) and \(b(x) = \pi\). This results in: \[y'=\left(-\frac{1+\sin ^{2} \pi}{1+\cos ^{2} \pi}\right)\left(-1\right)-\left(-\frac{1+\sin ^{2}(\pi-x)}{1+\cos^{2}(\pi-x)}\right)\left(-1\right)\]

Simplify Using Trigonometric Identities

Simplify the derivative using trigonometric identities. Since \(\sin(\pi)=0\) and \(\cos(\pi)=-1\), the first term simplifies to \(0\). Also, express \(\sin^{2}(\pi-x) = 1- \cos^{2}(\pi - x)\) and \(\cos(\pi - x) = -\cos(x)\). Substitute and simplify to get: \[y'=-\frac{1 +( 1 - \cos^{2}x)}{1 + \cos^{2}x}\] Simplifying further gives: \[y' = -\frac{2 - \cos^{2}x}{1 + \cos^{2}x}\]

Key concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a critical link between the operations of differentiation and integration in calculus. It bridges these two core concepts by stating that if we have a continuous function over an interval and we integrate it, then differentiate it, we essentially reverse the process and regain the original function.

In simple terms, if you have a function that is integrated from point 'a' to point 'b', the derivative of this integral, with respect to 'x', will be the value of the original function at 'x'. This theorem is comprised of two parts:

• Part 1 establishes the existence of antiderivatives for continuous functions.
• Part 2, often used in solving problems like the exercise provided, deals with evaluating the derivative of an integral as described above.

When working through textbook problems involving the Fundamental Theorem of Calculus, students should remember that it provides a straight-forward way to calculate the derivative of an integral, especially when the limits of integration are functions of the variable you're differentiating with respect to.

Integration

Integration is the process of finding the integral of a function, which represents the accumulation of quantities, such as area under a curve. In calculus education, it's generally taught after students have a firm grasp of differentiation, because it is, in a sense, the inverse operation.

Integrals can be of two types: indefinite and definite. An indefinite integral, also known as an antiderivative, represents a family of functions and includes a constant of integration. A definite integral, on the other hand, represents a number and is evaluated over a specified interval. The problem from the textbook requires finding a definite integral over the interval from \(\pi\) to \(\pi-x\), which represents a numeric area.

When evaluating definite integrals, students are often required to manipulate the limits of integration, as well as apply various integral properties and techniques to find the solution.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are true for all values of the variables involved. These identities are essential tools for simplifying and evaluating expressions in both trigonometry and calculus. Some of the most fundamental identities are Pythagorean identities, like \(\sin^2(x) + \cos^2(x) = 1\), and the angle sum and difference identities.

In the context of the given exercise, knowing these identities allows students to simplify the integrand and the resulting derivative in a meaningful way. For instance, recognizing that \(\sin^2(\pi-x) = 1 - \cos^2(\pi-x)\) is a consequence of the Pythagorean identity, and being familiar with the fact that \(\cos(\pi - x) = -\cos(x)\) are crucial steps in simplifying the derivative of the integral accordingly.

Calculus Education

Calculus education aims to develop students' understanding of concepts like limits, derivatives, integrals, and the connections between them. These topics are critical for solving problems in physics, engineering, economics, and beyond. A strong foundation in these areas enables students to approach complex, multi-step calculations with confidence.

In the exercise at hand, the educator’s goal is to guide students logically through the problem using a step-by-step approach that clarifies why each step is taken. It's not just about reaching the answer, but understanding the process: reversing the limits of integration, applying the Fundamental Theorem of Calculus, and using trigonometric identities to simplify the result. This approach fosters deeper comprehension and equips students with the skills to tackle a variety of calculus problems in the future.