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Evaluate the integrals. Some integrals do not require integration by parts. $$ \int \frac{\cos \sqrt{x}}{\sqrt{x}} d x $$

Short Answer

Expert verified
The integral evaluates to \( 2 \sin(\sqrt{x}) + C \).

Step by step solution

01

Identify the type of integral

The integral \( \int \frac{\cos \sqrt{x}}{\sqrt{x}} \, dx \) involves a composition of functions where the outer function is cosine and the inner function is a root function. This suggests a substitution might simplify the integral.
02

Apply u-substitution

Let \( u = \sqrt{x} \). Then \( du = \frac{1}{2\sqrt{x}} \, dx \) or \( 2u \, du = \, dx \). Substitute these into the integral, giving \( \int \frac{\cos u}{u} \, (2u \, du) \), which simplifies to \( 2 \int \cos u \, du \).
03

Integrate with respect to u

The integral \( 2 \int \cos u \, du \) is straightforward. The antiderivative of \( \cos u \) is \( \sin u \), so we integrate to get \( 2 \sin u + C \).
04

Substitute back to x

Finally, substitute \( u = \sqrt{x} \) back into the expression, yielding \( 2 \sin(\sqrt{x}) + C \) as the result.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

u-substitution
U-substitution is a clever integration technique that involves changing the variable of integration. This transformation simplifies the integration of complex expressions. In this case, we used it to change the form of the integral, making it easier to solve.

Imagine you're driving on a winding road. U-substitution helps by paving a smoother path, making your journey straightforward. The goal is to identify a part of the integrand that can be represented as a single variable, "u."

To guide you, look for compositions where one function is nested within another. Here, - Set \( u \) as the inner function, in this case, \( \sqrt{x} \).- Calculate \( du \), the derivative of \( u \), which represents the change in \( u \) as \( x \) changes.

Replace all occurrences of the original variables and their differentials with these new terms. This redefines the integrand, often into a product of simpler functions, leading to an easier integration process.
antiderivatives
Antiderivatives are essential in integration, as they reverse the differentiation process. Just like solving a puzzle by putting the pieces back in place, integrating involves finding the original function from its derivative.

When performing integration, finding antiderivatives is at the heart of the action. If you have the derivative \( \frac{d}{dx}f(x) \), then its antiderivative is simply \( f(x) \).

For instance, consider the function \( \cos u \). Its antiderivative is \( \sin u \). Thus, integrating \( \cos u \) gives you \( \sin u \). - Recall that when integrating, you add a constant \( C \) because differentiation of constants yields zero.

This constant ensures all possible vertical shifts of the function are accounted for. In simple terms, always remember to "add \( C \) to see" all potential solutions when integrating.
composite functions
Composite functions involve one function wrapped around another. Imagine a gift box nested inside another larger box. Similarly, the function \( f(g(x)) \) contains an outer function \( f \) and an inner function \( g \).

In integration, identifying composite functions helps streamline the problem-solving process. They often indicate where to use u-substitution.

For the given integral, the outer function is \( \cos \, u \) and the inner is the root function \( u = \sqrt{x} \).
  • The outer function dictates the larger scale behavior of the composite.
  • The inner function determines the specific path the outer function follows.
By understanding these roles, you can effectively manipulate and solve integrals involving compositions, like peeling layers off an onion.

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