Question

In the following exercises, find each indefinite integral by using appropriate substitutions. $$ \int e^{\sin x} \cos x d x $$

Short answer

The indefinite integral is \( e^{\sin x} + C \).

Step-by-step solution

Identify the Substitution

To solve the integral \( \int e^{\sin x} \cos x \; dx \), we notice a function and its derivative present. We select a substitution that simplifies the expression. Here, we set \( u = \sin x \), because its derivative, \( \frac{du}{dx} = \cos x \), is also part of the integrand.

Compute the Differential

Differentiate the substitution equation \( u = \sin x \) to find \( du \): \[du = \cos x \; dx\] This allows us to replace \( \cos x \; dx \) in the integral with \( du \).

Rewrite the Integral

Substitute the expressions for \( u \) and \( du \) into the integral: \[\int e^{\sin x} \cos x \; dx = \int e^u \; du\]The integral now takes the simpler form \( \int e^u \; du \).

Integrate with Respect to u

Integrate the simpler expression \( \int e^u \; du \): \[\int e^u \; du = e^u + C\] where \( C \) is the constant of integration.

Back-Substitute to x

Replace \( u \) with the original substitution \( \sin x \) to get back to the variable \( x \): \[e^{u} + C = e^{\sin x} + C\] Thus, the indefinite integral is \( e^{\sin x} + C \).

Key concepts

Substitution Method

The substitution method is a powerful technique used in calculus to simplify the process of integration. It's often applied when an integral contains a composite function where the inner function's derivative is also present in the integrand. In our context, we faced the integral \( \int e^{\sin x} \cos x \, dx \).

• First, we identify a substitution. A good choice is to set \( u = \sin x \), given that the derivative \( du = \cos x \, dx \) is readily available in the integral.
• Next, rewrite the original integral using the substitution: \( u \) replaces \( \sin x \) and \( du \) replaces \( \cos x \, dx \). This transition simplified our integral to \( \int e^u \, du \).

The substitution method effectively reduces the complexity of the integral, transforming it into a more straightforward standard form that is often easier to solve. In this case, the exponential function of \( e^u \) is one of the simplest forms to integrate.

Differential Calculus

Differential calculus plays a crucial role in the substitution method, especially in computing derivatives. It involves finding derivatives which represent the rate at which one quantity changes with respect to another. Here’s how it applies:

• When we set \( u = \sin x \), differential calculus helps us compute \( du \), the differential that replaces \( \cos x \, dx \) in the integral.
• The derivative of \( \sin x \) with respect to \( x \) is \( \cos x \), hence \( du = \cos x \, dx \).

Understanding how to differentiate effectively is critical. It allows us to adjust the integral expression so it can transition smoothly into a solvable form. Without derivative computation, finding the correct substitutions would be challenging and potentially incorrect. Differential calculus thus provides the foundation for successfully employing substitution.

Exponential Functions

Exponential functions appear in various mathematical contexts and are characterized by their constant base raised to a variable exponent. In integration, understanding their properties can simplify computations greatly. In our exercise, the function \( e^{\sin x} \) is present.

• By substituting properly, the function converts into \( e^u \), which is an exponential function with base \( e \), the natural exponential base, and exponent \( u \).
• The integration of the form \( \int e^u \, du \) is particularly straightforward: it is \( e^u + C \).

This integral forms a basic rule in calculus, where the derivative and the integral of \( e^x \) are both \( e^x \). Thus, once the simplified integral appears, it can be directly computed without further breakdown or complex integration techniques. Understanding exponential functions is vital for quick recognition and solution of such integrals.