Problem 16 - Biocalculus Calculus for the Life Sciences

Question

Evaluate the indefinite integral. \(\int \frac{\sin \sqrt{x}}{\sqrt{x}} d x\)

Step-by-Step Solution

Answer

The indefinite integral is \(-2\cos \sqrt{x} + C\).

Step 1: Recognize the Need for Substitution

To integrate the expression \( \int \frac{\sin \sqrt{x}}{\sqrt{x}} \, dx \), notice that \( \sqrt{x} \) appears in both the sine function and the denominator. This suggests the possibility of using a substitution that simplifies the expression.

Step 2: Define the Substitution

Let \( u = \sqrt{x} \). Then, \( u^2 = x \) and by differentiating both sides, we get \( 2u \, du = dx \). Thus, \( dx = 2u \, du \). This substitution will simplify the integral.

Step 3: Substitute and Simplify the Integral

Substitute \( x = u^2 \) and \( dx = 2u \, du \) into the integral. The expression becomes \( \int \frac{\sin u}{u} \, 2u \, du \). Simplify this expression to \( 2 \int \sin u \, du \).

Step 4: Integrate the Simplified Function

The integral \( \int \sin u \, du \) is a standard integral. Its antiderivative is \(-\cos u + C\), where \( C \) is the constant of integration. Therefore, \( 2 \int \sin u \, du = -2\cos u + C\).

Step 5: Back-Substitute the Original Variable

Replace \( u \) with the original variable \( \sqrt{x} \) as per the substitution \( u = \sqrt{x} \). This means the antiderivative is \(-2\cos \sqrt{x} + C\).

Key Concepts

Substitution Method

In calculus, the substitution method is a powerful tool for solving indefinite integrals. It involves replacing a complicated part of the integrand with a new variable to make the integration process simpler. In our example, we have the integral \( \int \frac{\sin \sqrt{x}}{\sqrt{x}} \, dx \). The substitution method begins by identifying a part of the expression that, when substituted, will simplify the integration. Here, \( \sqrt{x} \) is a fitting candidate because it appears both in the sine function and as its own denominator.
To apply substitution, you set \( u = \sqrt{x} \). Differentiating \( u = \sqrt{x} \) gives \( 2u \, du = dx \). This is essential for rewriting the integral in terms of \( u \), thereby changing the problem into something easier to handle.

Integration Techniques

Integration is a method for finding a function's antiderivative, and there are several techniques to achieve this, including substitution and integration by parts. In this exercise, substitution is the most appropriate technique. The goal when using any integration technique is to convert the integral into a form that is easily recognizable or solves into a basic known integral.
By substituting \( x = u^2 \) and \( dx = 2u \, du \) into the integral, the original complex integral \( \int \frac{\sin \sqrt{x}}{\sqrt{x}} \, dx \) simplifies to a much more manageable \( 2 \int \sin u \, du \). This new expression is in a standard form that can be integrated directly.

Antiderivative

The antiderivative, or indefinite integral, of a function represents a family of functions whose derivative is the original function. In our integral, once we have transformed it via substitution into \( 2 \int \sin u \, du \), we integrate to find the antiderivative of \( \sin u \).
The antiderivative of \( \sin u \) is \(-\cos u\), so the antiderivative of our transformed integral becomes \(-2\cos u + C\), where \( C \) is the constant of integration. This constant reflects the fact that there are infinitely many antiderivatives for a function, each differing by a constant. Finally, you must reverse the substitution, replacing \( u \) with \( \sqrt{x} \), giving the complete antiderivative: \(-2\cos \sqrt{x} + C\).

Trigonometric Integrals

Trigonometric integrals involve trigonometric functions, like sine and cosine, which are common in calculus. These types of integrals can often be simplified using substitution or trigonometric identities to make integration feasible. Integrals involving trigonometric functions can frequently be evaluated by recalling the basic antiderivatives of sine and cosine.
In the exercise, once substituted, our integral \( \int \sin u \, du \) was easily recognized as a standard trigonometric integral. Knowing the basic integration rules, such as \( \int \sin u \, du = -\cos u + C\), allows us to quickly find solutions. Trigonometric integrals often require practice and familiarity with fundamental trigonometric identities and integration rules to efficiently solve them.