Chapter 6: Problem 57
Evaluate the indefinite integral. \(\int \sin (x) \sqrt{\cos (x)} d x\)
Short Answer
Expert verified
The integral is \(-\frac{2}{3} (\cos(x))^{3/2} + C\).
Step by step solution
01
Select a Suitable Substitution
To evaluate the integral \(\int \sin(x) \sqrt{\cos(x)} \, dx\), we can use substitution. Notice the derivative of \(\cos(x)\) is \(-\sin(x)\), which suggests setting \(u = \cos(x)\). This simplifies expressions related to \(\cos(x)\) and \(\sin(x)\) into terms of \(u\).
02
Differentiate the Substitution
Differentiate \(u = \cos(x)\) to find \(du\). This gives \(du = -\sin(x) \, dx\), or \(-du = \sin(x) \, dx\). We will use this substitution in the integral.
03
Substitute in the Integral
Replace \(\sin(x) \, dx\) with \(-du\) and \(\sqrt{\cos(x)}\) with \(\sqrt{u}\) in the integral. The integral becomes \(-\int \sqrt{u} \, du\).
04
Integrate the New Expression
Evaluate the integral \(-\int \sqrt{u} \, du\). Rewrite \(\sqrt{u}\) as \(u^{1/2}\). The integral is now \(-\int u^{1/2} \, du\). Using the power rule of integration, this becomes:\[-\left(\frac{2}{3}u^{3/2}\right) + C\], where \(C\) is the constant of integration.
05
Substitute Back to Original Variables
Replace \(u\) back with \(\cos(x)\), the original variable. Thus, the integral in terms of \(x\) is:\[-\frac{2}{3} (\cos(x))^{3/2} + C\].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Indefinite Integral
An indefinite integral is a fundamental concept in calculus that represents the collection of all antiderivatives of a function. When you compute an indefinite integral, you are essentially reversing the process of differentiation. It's like asking, "What function, when differentiated, gives us this function?" The indefinite integral of a function is usually expressed with the integral sign and a constant of integration.
- The integral sign \( \int \) indicates the operation of integration.
- The function to be integrated is known as the integrand.
- The result is a general form including a constant, denoted by \( C \), since differentiation of a constant is zero.
Substitution Method
The substitution method is a powerful technique in integral calculus used to simplify integrals. It's especially handy when the integrand is a composite function. The essence of this method is to change variables such that the integral becomes easier to evaluate.
- Choose substitution wisely: Look for functions whose derivatives are present in the integrand.
- Relate \( du \) back to \( dx \): Differentiate the substitution to find the differential change.
- Simplify and integrate: Rewrite the integral with the new variable for a simpler form.
Power Rule of Integration
The power rule of integration is a straightforward and essential tool for antiderivatives. It says that when integrating a power function, you increase the exponent by one, and then divide by this new exponent.
Here's the rule for any function \( x^n \):
In our integral, upon substituting, we have \[ -\int u^{1/2} \, du, \] which follows directly from the power rule. We add one to the exponent \( 1/2 \), making it \( 3/2 \), and then divide by the new exponent:\[ -\left(\frac{2}{3} u^{3/2}\right) + C. \] This calculation is straightforward yet foundational, demonstrating how the power rule enables us to find antiderivatives of polynomials efficiently.
Here's the rule for any function \( x^n \):
- \( \int x^n \, dx = \frac{1}{n+1} x^{n+1} + C, \quad n eq -1, \)
In our integral, upon substituting, we have \[ -\int u^{1/2} \, du, \] which follows directly from the power rule. We add one to the exponent \( 1/2 \), making it \( 3/2 \), and then divide by the new exponent:\[ -\left(\frac{2}{3} u^{3/2}\right) + C. \] This calculation is straightforward yet foundational, demonstrating how the power rule enables us to find antiderivatives of polynomials efficiently.