Question

Given the integral \(\int_{0}^{\pi} \frac{d x}{1+\cos ^{2} x}\). Make sure that the functions \(\mathrm{F}_{1}(\mathrm{x})=\frac{1}{\sqrt{2}} \cos ^{-1} \frac{\sqrt{2} \cos \mathrm{x}}{\sqrt{1+\cos ^{2} \mathrm{x}}}\) and \(\mathrm{F}_{2}(\mathrm{x})=\frac{1}{\sqrt{2}} \tan ^{-1} \frac{\tan \mathrm{x}}{\sqrt{2}}\) are antiderivatives for the integrand. Is it possible to use both antiderivatives for computing the definite integral by the NewtonLeibnitz formula? If not, which of the antiderivatives can be used?

Short answer

Answer: Yes, both antiderivatives \(F_1(x)\) and \(F_2(x)\) can be used to compute the definite integral using the Newton-Leibnitz formula.

Step-by-step solution

Verify the antiderivatives

To verify if the functions are antiderivatives, we need to find the derivatives of \(F_1(x)\) and \(F_2(x)\) and see if they match with the integrand function. We'll start with \(F_1(x)\). $$F_1(x)=\frac{1}{\sqrt{2}} \cos ^{-1} \frac{\sqrt{2} \cos x}{\sqrt{1+\cos ^{2} x}}$$ Now, we differentiate \(F_1(x)\) with respect to \(x\): $$F_1'(x) = \frac{d}{dx}\left(\frac{1}{\sqrt{2}} \cos ^{-1} \frac{\sqrt{2} \cos x}{\sqrt{1+\cos ^{2} x}}\right)$$ By using the chain rule, we find the derivative to be: $$F_1'(x) = \frac{1}{1 + \cos^2 x}$$ Now, let's differentiate \(F_2(x)\): $$F_2(x) = \frac{1}{\sqrt{2}} \tan ^{-1} \frac{\tan x}{\sqrt{2}}$$ Now, we differentiate \(F_2(x)\) with respect to \(x\): $$F_2'(x) = \frac{d}{dx}\left(\frac{1}{\sqrt{2}} \tan ^{-1} \frac{\tan x}{\sqrt{2}}\right)$$ By using the chain rule, we find the derivative to be: $$F_2'(x) = \frac{1}{1 + \cos^2 x}$$ Since both derivatives match the integrand function, both \(F_1(x)\) and \(F_2(x)\) are antiderivatives of \(\frac{1}{1+\cos^2 x}\).

Identify which antiderivative to use for Newton-Leibnitz formula

To determine which antiderivative can be used to compute the definite integral using the Newton-Leibnitz formula, we need to consider the domains of the functions \(F_1(x)\) and \(F_2(x)\). For \(F_1(x)\), the arccosine (\(\cos^{-1}(\cdot)\)) function is defined only when \(-1\leq \frac{\sqrt{2} \cos x}{\sqrt{1+\cos ^{2} x}}\leq1\). For \(F_2(x)\), the arctangent (\(\tan^{-1}(\cdot)\)) function is defined for all real values of \(x\). Since the given integral is between the limits 0 and \(\pi\), the function \(F_1(x)\) might have some issues with its domain. However, looking closer at the expression inside the arccosine function: $$-1\leq \frac{\sqrt{2} \cos x}{\sqrt{1+\cos ^{2} x}}\leq1$$ We find that the expression is always within the range of [-1, 1] for all \(x \in [0, \pi]\). Thus, both antiderivatives can be used for computing the definite integral using the Newton-Leibnitz formula. In conclusion, it's possible to use both antiderivatives (\(F_1(x)\) and \(F_2(x)\)) for computing the definite integral \(\int_{0}^{\pi} \frac{dx}{1+\cos ^{2} x}\) using the Newton-Leibnitz formula.

Key concepts

Newton-Leibnitz Formula

The Newton-Leibnitz formula is a fundamental theorem in integral calculus that connects antiderivatives with definite integrals. It states that if a function has an antiderivative over an interval, the definite integral of the function over that interval can be evaluated using the values of its antiderivative at the endpoints.

The formula is generally given as:
\[\begin{equation}\begin{aligned}\text{If } F(x) \text{ is an antiderivative of } f(x), \text{ then } \ \text{the definite integral over } [a, b] \text{ is given by: } \text{ } \text{ }\begin{aligned}\int_{a}^{b} f(x) \text{ d}x = F(b) - F(a).\end{aligned}\end{aligned}\end{equation}\]
For instance, in the given exercise, if we can verify the antiderivatives, we can then apply the Newton-Leibnitz formula to compute the definite integral from 0 to \(\pi\) for the integrand \(\frac{1}{1 + \cos^2 x}\). This process greatly simplifies solving complex integrals and is an essential tool for calculus students to master.

Antiderivatives

Antiderivatives, also known as indefinite integrals, are functions that describe the accumulation of the rate of change (given by derivatives) and thereby reverse differentiation. The search for antiderivatives is a central task in solving integral problems, as it allows one to use the Newton-Leibnitz formula to evaluate definite integrals.

Mathematically, an antiderivative for a function \(f(x)\) is another function \(F(x)\) such that:\[F'(x) = f(x).\]
Finding an antiderivative requires knowledge of fundamental integration techniques and often involves recognizing patterns or employing rules such as the chain rule to work with composite functions. While commonly known functions have established antiderivatives, unique cases might require creativity and a deep understanding of calculus principles.

Definite Integral

The definite integral is a core concept in calculus representing the accumulation or net area under a curve between two specified points. Unlike indefinite integrals, which yield a family of functions plus a constant (the antiderivative), a definite integral provides a specific numerical value representing this total accumulation.

Definite integrals have a wide range of applications, including calculating areas, volumes, and other physical quantities like work and center of mass. Executing a definite integral involves finding the appropriate antiderivative and using the Newton-Leibnitz formula to evaluate it between the given limits. Understanding how to transition from the abstract concept of net area to its practical computation is essential for solving calculus problems.

Chain Rule

The chain rule is a fundamental differentiation technique in calculus, used when taking the derivative of composite functions. When we have a function composed within another function, \(f(g(x))\), the chain rule tells us to differentiate the outer function, \(\f'\), and then multiply by the derivative of the inner function, \(\f(x)\).

The chain rule is vital for finding derivatives efficiently and accurately, especially when identifying antiderivatives as part of solving integral problems. As seen in the textbook exercise, the chain rule helped verify that both presented functions were valid antiderivatives of the integrand by differentiating them and showing that their derivatives match the original integrand function.