Problem 24 - AP Calculus with 8 Practice Tests 15th

Question

\(\int \frac{\sin \sqrt{x}}{\sqrt{x}} d x=\) (A) \(-2 \cos ^{1 / 2} x+C\) (B) \(-\cos \sqrt{x}+C\) (C) \(-2 \cos \sqrt{x}+C\) (D) \(\frac{1}{2} \cos \sqrt{x}+C\)

Step-by-Step Solution

Answer

The correct option is (C): -2 \cos \sqrt{x}+C.

Step 1: Use Substitution

To solve the integral \( \int \frac{\sin \sqrt{x}}{\sqrt{x}} \, dx \), we use the substitution method. Let \( u = \sqrt{x} \), then \( du = \frac{1}{2\sqrt{x}} \, dx \) or \( dx = 2u \, du \).

Step 2: Substitute in the Integral

Substitute \( u = \sqrt{x} \) and \( dx = 2u \, du \) into the integral: \( \int \frac{\sin u}{u} \, 2u \, du = 2 \int \sin u \, du \). The \( u \) terms cancel each other.

Step 3: Integrate the Substituted Integral

Now, integrate \( 2 \int \sin u \, du \). The integral of \( \sin u \) is \(-\cos u\), hence \( 2 \int \sin u \, du = -2\cos u + C \).

Step 4: Replace the Substitute Variable

Replace \( u \) back with \( \sqrt{x} \). So, the integral becomes \( -2\cos \sqrt{x} + C \).

Step 5: Identify the Correct Option

Compare the result \( -2\cos \sqrt{x} + C \) with the given options. This matches with option (C): \(-2 \cos \sqrt{x} + C\).

Key Concepts

Definite Integrals

Definite integrals are an essential concept in calculus. A definite integral is used to compute the accumulation of quantities, such as areas under curves. For example, it can calculate the area between a curve, the x-axis, and vertical lines at two different points on the x-axis. The notation for a definite integral is \( \int_a^b f(x) \, dx \). Here, \( a \) and \( b \) are the limits of integration.
These limits show the interval over which you calculate the integral. The result of a definite integral is a number, representing the net area, which can be positive, negative, or zero. Keep in mind the properties of definite integrals:

• If you reverse the limits, the sign of the integral changes.
• Integrals of functions with respect to different intervals can be split or combined.
• If the function is zero over an interval, the integral's value is zero.

Understanding these helps allow you to solve problems efficiently, especially when functions are complex.

Substitution Method

The substitution method, often called u-substitution, is a powerful technique to simplify and solve integrals. It's similar to reversing the chain rule for derivatives. Use this when an integral contains a composite function and you can identify a part of the function that serves as an inner layer.
For instance, consider the integral \( \int \frac{\sin \sqrt{x}}{\sqrt{x}} \, dx \). Here, take \( u = \sqrt{x} \) because it's the inner function under both sine and square root. Then find \( du \). It is effective because it allows changing variables to simplify the integral, often leading to a straightforward function that you can easily integrate.
Steps for using substitution:

• Select a substitution: let \( u = g(x) \).
• Find \( du = g'(x) \, dx \).
• Rewrite the integral in terms of \( u \).
• Integrate with respect to \( u \).
• Finally, substitute back the original variable.

The substitution method greatly facilitates solving integrals, making it easier to manage and integrate complex functions.

Integration Techniques

Integration techniques are essential for calculating integrals, which are a fundamental part of calculus. Apart from basic integration rules, techniques like substitution, integration by parts, and partial fractions allow for the integration of more complicated expressions.
Each method has its own scope and application. For example, substitution, like in our original problem, is best used when an integral has a composite function structure. Meanwhile, integration by parts applies well to products of polynomials and exponential or trigonometric functions.
Here's a brief overview of common methods:

• Substitution: Simplifies the integral by changing variables.
• Integration by Parts: Based on the product rule for derivatives, expressed as \( \int u \, dv = uv - \int v \, du \).
• Partial Fraction Decomposition: Breaks a rational function into simpler fractions that are easier to integrate.
• Trigonometric Identities: Utilizes identities such as \( \sin^2 x + \cos^2 x = 1 \) for integration.

Proficiency in selecting the correct method enhances the ability to tackle diverse integration problems with confidence.