Problem 39
Question
First make an appropriate substitution and then use integration by parts to evaluate the indefinite integrals. $$ \int \cos \sqrt{x} d x $$
Step-by-Step Solution
Verified Answer
The integral evaluates to \( 2 \sqrt{x} \sin(\sqrt{x}) + 2 \cos(\sqrt{x}) + C \).
1Step 1: Substitution
To simplify the integral, we make a substitution. Let \( u = \sqrt{x} \), which implies \( x = u^2 \). The differential \( dx \) changes as follows: \( dx = 2u \, du \). Substituting these into the integral gives:\[\int \cos(u) \cdot 2u \, du = 2 \int u \cos(u) \, du\]
2Step 2: Set Up Integration by Parts
To apply integration by parts, we choose functions for \( u \) and \( dv \). Let \( v = u \) and \( dw = \cos(u) \, du \). The integration by parts formula is:\[\int u \, dv = uv - \int v \, du\]We have:\[du = 1 \, du, \quad v = \int \cos(u) \, du = \sin(u)\]
3Step 3: Apply Integration by Parts
Substituting \( v \) and \( du \) into the formula:\[2 \int u \cos(u) \, du = 2 \times \left( u \sin(u) - \int \sin(u) \, du \right)\]Evaluating the remaining integral \( \int \sin(u) \, du \) gives \(-\cos(u) + C_1\). Thus:\[2 \times \left( u \sin(u) + \cos(u) \right)\]
4Step 4: Substitute Back Original Variable
Return to the original variable \( x \) by replacing \( u \) with \( \sqrt{x} \) in the result:\[2 \cdot ( \sqrt{x} \sin(\sqrt{x}) + \cos(\sqrt{x}) ) + C\]Thus, the indefinite integral evaluates to:\[2 \sqrt{x} \sin(\sqrt{x}) + 2 \cos(\sqrt{x}) + C\]
Key Concepts
Substitution MethodIndefinite IntegralsDifferential Calculus
Substitution Method
The substitution method is a powerful technique in calculus used to simplify the process of integration. It revolves around replacing a complicated expression with a single variable to make integration easier.
In our example, we are dealing with the integral \( \int \cos \sqrt{x} \, dx \). By substituting \( u = \sqrt{x} \), we transform the integral into a more manageable form. This changes our original variable \( x \) into \( u^2 \), which also means that \( dx \) is replaced by \( 2u \, du \).
Now, instead of integrating with respect to \( x \), you're integrating with respect to \( u \), giving us \( 2 \int u \cos(u) \, du \). The substitution simplifies the process and is critical for tackling more complex integrals. Remember, the aim is to make the function easier or reduce a complex integral into a standard form that we know how to integrate directly.
In our example, we are dealing with the integral \( \int \cos \sqrt{x} \, dx \). By substituting \( u = \sqrt{x} \), we transform the integral into a more manageable form. This changes our original variable \( x \) into \( u^2 \), which also means that \( dx \) is replaced by \( 2u \, du \).
Now, instead of integrating with respect to \( x \), you're integrating with respect to \( u \), giving us \( 2 \int u \cos(u) \, du \). The substitution simplifies the process and is critical for tackling more complex integrals. Remember, the aim is to make the function easier or reduce a complex integral into a standard form that we know how to integrate directly.
Indefinite Integrals
Indefinite integrals represent a family of functions whose derivative gives the original function you have integrated. They are usually denoted without upper and lower limits of integration, represented simply as \( \int f(x) \, dx \).
These integrals are distinguished by the inclusion of a constant \( C \) in their solution. This constant arises because the differentiation of a constant is zero, so it doesn't affect the derivative.
The purpose of the indefinite integral in calculus is to find antiderivatives. When you integrate \( \int \cos \sqrt{x} \, dx \), you're looking for a function whose derivative will return the original function, plus the constant of integration \( C \) to account for all possible antiderivatives.
These integrals are distinguished by the inclusion of a constant \( C \) in their solution. This constant arises because the differentiation of a constant is zero, so it doesn't affect the derivative.
The purpose of the indefinite integral in calculus is to find antiderivatives. When you integrate \( \int \cos \sqrt{x} \, dx \), you're looking for a function whose derivative will return the original function, plus the constant of integration \( C \) to account for all possible antiderivatives.
Differential Calculus
Differential calculus focuses on the concept of the derivative, which measures how a function changes as its input changes. It is the study of how variables change in relation to one another and is foundational in calculus.
In the original exercise, the differential element \( dx \) transitions to \( 2u \, du \) due to substitution, following the rules of differential calculus. This is calculated by differentiating \( x = u^2 \), which leads us to \( dx = 2u \, du \).
This differential aspect is key to adjusting variables and is crucial throughout calculus to understand changes in functions. Differential calculus is also fundamental when applying methods like integration by parts, since integration and differentiation are inverse operations that work hand-in-hand.
In the original exercise, the differential element \( dx \) transitions to \( 2u \, du \) due to substitution, following the rules of differential calculus. This is calculated by differentiating \( x = u^2 \), which leads us to \( dx = 2u \, du \).
This differential aspect is key to adjusting variables and is crucial throughout calculus to understand changes in functions. Differential calculus is also fundamental when applying methods like integration by parts, since integration and differentiation are inverse operations that work hand-in-hand.
Other exercises in this chapter
Problem 38
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