Question

Integrating factor of differential equation \([1 /(\cos x)] \cdot(d y / d x)+[1 /(\sin x)] y=1\) is. (A) \(\sec x\) (B) \(\cos \mathrm{x}\) (C) \(\tan \mathrm{x}\) (D) \(\sin \mathrm{x}\)

Short answer

The integrating factor of the given differential equation is \(\tan x\) (C).

Step-by-step solution

Identify P(x) and Q(x)

Given the differential equation \(\frac{1}{\cos x} \frac{dy}{dx} + \frac{1}{\sin x}y = 1\), we can see that it is in the form of a first-order linear differential equation with P(x) = \(\frac{1}{\sin x}\) and Q(x) = 1.

Find the integrating factor

As mentioned earlier, the integrating factor is given by the formula \(e^{\int P(x)dx}\). Let's find \(\int P(x)dx\): \[\int P(x)dx = \int \frac{1}{\sin x} dx\] Let's make a substitution: \(u = \sin x, du = \cos x dx\) \[\int \frac{1}{u} \cdot \frac{du}{\cos x} = \frac{1}{\cos x}\int \frac{1}{u} du\] Now let's integrate: \[ \frac{1}{\cos x}\int \frac{1}{u} du = \frac{ln|u|+c}{\cos x} = \frac{ln|\sin x| + c}{\cos x}\] Now, let's find the integrating factor: \[e^{\int P(x)dx} = e^{\frac{ln|\sin x| + c}{\cos x}}\] But we know that \(e^{\ln z} = z\), so: \[e^{\int P(x)dx} = \frac{e^{ln|\sin x|+c}}{\cos x} = \frac{\sin x}{\cos x}\] Thus, the integrating factor is \(\tan x\), which corresponds to the answer choice (C).

Key concepts

First-Order Linear Differential Equations

A first-order linear differential equation involves a derivative and a linear combination of the function and its derivative. In general, this equation can be written in the form \( \frac{dy}{dx} + P(x)y = Q(x) \). Here, the term \( \frac{dy}{dx} \) represents the rate of change of a function \( y \), and both \( P(x) \) and \( Q(x) \) are functions of \( x \).
Often students encounter these equations in calculus, as they lay the foundation for understanding how systems change over time. They are widely used in physics, biology, and economics, for example, to model exponential growth or decay problems.

• P(x): This is the coefficient function of \( y \).
• Q(x): This is the function on the right-hand side of the differential equation.

For instance, in the given differential equation \( \frac{1}{\cos x} \frac{dy}{dx} + \frac{1}{\sin x} y = 1 \), applying this structure, \( P(x) \) becomes \( \frac{1}{\sin x} \) and \( Q(x) \) is \( 1 \). Understanding these components is crucial when you begin solving these differential equations because they guide you on applying the integrating factor.

Integration Techniques

Integration is a crucial technique in calculus used to solve differential equations, among other applications. In our problem, we use integration to find the integrating factor, a function that simplifies the process of solving the differential equation.
When integrating a function such as \( \int \frac{1}{\sin x} dx \), our goal is to transform it into a more manageable form. This often involves using substitution methods or recognizing it as a standard integral form.
One common integration technique involves recognizing parts of the integrand can be substituted to make the integral simpler. For example:

• Consider the integral \( \int \frac{1}{\sin x} dx \).
• We substitute \( u = \sin x \) making \( du = \cos x dx \).
• This substitution transforms our integral into \( \int \frac{1}{u} du \), a standard form whose solution is \( \ln|u| + c \).
• Replacing \( u \) back with \( \sin x \), we get \( \ln|\sin x| + c \).

This process exemplifies how integration transforms challenging problems into solvable forms. It's a key process in calculus, aiding in solving not only differential equations but also evaluating area, volume, and more.

Substitution Method in Calculus

The substitution method, often referred to as "u-substitution," is an essential tool for simplifying integrals in calculus. It essentially helps convert a complicated integral into a simpler form that is easier to evaluate.
In solving the given exercise, we are tasked with integrating \( \int \frac{1}{\sin x} dx \). Direct integration is tough, so we employ substitution:

• First, identify a substitution that simplifies the integral. Here, let \( u = \sin x \).
• Then, find \( du \), the derivative of \( u \), which equals \( \cos x dx \).
• Rewrite the integral in terms of \( u \): it becomes \( \int \frac{1}{u} du \).
• The transformed integral \( \int \frac{1}{u} du \) is simpler and leads to the result \( \ln|u| + c \).
• Substitute \( u \) back to get the final result in terms of \( x \), resulting in \( \ln|\sin x| + c \).

Using substitution not only helps in representing the function in a solvable manner but also enhances understanding of the integral's behavior. Mastery of this technique is invaluable for tackling more complex calculus tasks.