Problem 15
Question
In Problems 1-54, perform the indicated integrations. \(\int_{0}^{\pi / 4} \frac{\cos x}{1+\sin ^{2} x} d x\)
Step-by-Step Solution
Verified Answer
The integral evaluates to \( \frac{\pi}{8} \).
1Step 1: Understand the Problem
We are tasked with evaluating the definite integral \( \int_{0}^{\pi / 4} \frac{\cos x}{1+\sin ^{2} x} \, dx \). This involves finding the area under the curve of the function \( \frac{\cos x}{1+\sin ^{2} x} \) from 0 to \( \frac{\pi}{4} \).
2Step 2: Simplify the Integrand
To simplify the problem, consider if any substitution might simplify the expression. Notice that the derivative of \( \sin x \) is \( \cos x \), suggesting a substitution strategy. Let \( u = \sin x \), then \( du = \cos x \, dx \). The limits of integration change correspondingly: when \( x = 0, \ u = 0 \) and when \( x = \frac{\pi}{4}, \ u = \frac{1}{\sqrt{2}} \).
3Step 3: Perform the Substitution
After substituting, the integral becomes: \( \int_{0}^{1/\sqrt{2}} \frac{1}{1+u^2} \, du \). The integrand \( \frac{1}{1+u^2} \) is a standard arctan integration form.
4Step 4: Integrate Using Standard Formulas
Recognize the integral \( \int \frac{1}{1+u^2} \, du = \arctan u + C \). Apply the indefinite integral formula to our definite integral: \( \int_{0}^{1/\sqrt{2}} \frac{1}{1+u^2} \, du = [\arctan u]_{0}^{1/\sqrt{2}} \).
5Step 5: Evaluate the Definite Integral
Calculate \( [\arctan u]_{0}^{1/\sqrt{2}} = \arctan\left(\frac{1}{\sqrt{2}}\right) - \arctan(0) \). Note that \( \arctan(0) = 0 \). Thus, the integral evaluates to \( \arctan\left(\frac{1}{\sqrt{2}}\right) \).
6Step 6: Find the Specific Value
Since \( \arctan\left(\frac{1}{\sqrt{2}}\right) = \frac{\pi}{8} \) due to special angle knowledge, our integral evaluates to \( \frac{\pi}{8} \).
Key Concepts
Definite IntegralTrigonometric SubstitutionArctan IntegrationIntegration Limits
Definite Integral
A definite integral represents the 'total accumulation' of a function within specified limits of integration. For the problem at hand, we're tasked with finding the definite integral of the function \(\frac{\cos x}{1+\sin ^{2} x}\) within the range from \(0\) to \(\frac{\pi}{4}\).
A definite integral can be visualized as the area under the curve of the function between these two points. Think of it as filling up a region with slices from 0 to \(\frac{\pi}{4}\) on the x-axis.
Calculating the definite integral involves integration techniques that simplify the integrand, allowing us to evaluate the total area effectively.
Always remember the limits, they define the start and end of the region of interest in a definite integral.
If you change them, you are essentially changing the problem that you are solving.
A definite integral can be visualized as the area under the curve of the function between these two points. Think of it as filling up a region with slices from 0 to \(\frac{\pi}{4}\) on the x-axis.
Calculating the definite integral involves integration techniques that simplify the integrand, allowing us to evaluate the total area effectively.
Always remember the limits, they define the start and end of the region of interest in a definite integral.
If you change them, you are essentially changing the problem that you are solving.
Trigonometric Substitution
Trigonometric substitution is a technique used to simplify integrals by substituting trigonometric identities. For the integral \(\int_{0}^{\pi / 4} \frac{\cos x}{1+\sin ^{2} x} \, dx\), we use trigonometric substitution to transform the function into a simpler form.
The function \(\sin x\) has a derivative \(\cos x\), making it a promising candidate for substitution.
The substitution we make is \(u = \sin x\), which implies that \(du = \cos x \, dx\).
This process simplifies our integrand to \(\frac{1}{1+u^2}\), a form known for further simplification with arctan integration.
This substitution also changes the integration limits, which are essential for recalculating the span of integration for our new variable.
The function \(\sin x\) has a derivative \(\cos x\), making it a promising candidate for substitution.
The substitution we make is \(u = \sin x\), which implies that \(du = \cos x \, dx\).
This process simplifies our integrand to \(\frac{1}{1+u^2}\), a form known for further simplification with arctan integration.
This substitution also changes the integration limits, which are essential for recalculating the span of integration for our new variable.
Arctan Integration
After substitution, our integral becomes \(\int_{0}^{1/\sqrt{2}} \frac{1}{1+u^2} \, du\). This is a standard form that simplifies to arctan integration.
A standard integration formula you should remember is \(\int \frac{1}{1+u^2} \, du = \arctan u + C\), where \(C\) is the constant of integration for indefinite integrals.
Because our problem is a definite integral, the constant \(C\) does not appear in the final evaluation.
The definite integral solution becomes \([\arctan u]_{0}^{1/\sqrt{2}}\). This means we need to calculate the arctan values at our new limits of integration, \(0\) and \(1/\sqrt{2}\).
Understanding and recognizing such standard forms helps streamline the integration process.
A standard integration formula you should remember is \(\int \frac{1}{1+u^2} \, du = \arctan u + C\), where \(C\) is the constant of integration for indefinite integrals.
Because our problem is a definite integral, the constant \(C\) does not appear in the final evaluation.
The definite integral solution becomes \([\arctan u]_{0}^{1/\sqrt{2}}\). This means we need to calculate the arctan values at our new limits of integration, \(0\) and \(1/\sqrt{2}\).
Understanding and recognizing such standard forms helps streamline the integration process.
Integration Limits
Integration limits define the boundaries of the region over which the definite integral is to be evaluated. In our problem, the original limits are \(0\) and \(\frac{\pi}{4}\).
However, once we make our substitution, these limits need to be updated accordingly.
For \(x = 0\), \(u = \sin(0) = 0\). Similarly, for \(x = \frac{\pi}{4}\), \(u = \sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}\).
This transformation ensures the integral remains accurate in the \(u\)-domain.
Correctly adjusting limits with a substitution is crucial to solving the problem correctly.
Finally, these adjusted limits are used in our arctan integration: \(\arctan\left(\frac{1}{\sqrt{2}}\right) - \arctan(0)\).
Grasping the significance of integration limits enhances the understanding and computation of definite integrals.
However, once we make our substitution, these limits need to be updated accordingly.
For \(x = 0\), \(u = \sin(0) = 0\). Similarly, for \(x = \frac{\pi}{4}\), \(u = \sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}\).
This transformation ensures the integral remains accurate in the \(u\)-domain.
Correctly adjusting limits with a substitution is crucial to solving the problem correctly.
Finally, these adjusted limits are used in our arctan integration: \(\arctan\left(\frac{1}{\sqrt{2}}\right) - \arctan(0)\).
Grasping the significance of integration limits enhances the understanding and computation of definite integrals.
Other exercises in this chapter
Problem 14
In Problems 1-14, solve each differential equation. $$ \sin x \frac{d y}{d x}+2 y \cos x=\sin 2 x ; y=2 \text { when } x=\frac{\pi}{6} $$
View solution Problem 14
In Problems 1-36, use integration by parts to evaluate each integral. $$ \int \arctan 5 x d x $$
View solution Problem 15
In Problems 11-16, use Euler's Method with \(h=0.2\) to approximate the solution over the indicated interval. $$ y^{\prime}=x y, y(1)=1,[1,2] $$
View solution Problem 15
In Problems 1-28, perform the indicated integrations. \(\int \sin ^{4}\left(\frac{w}{2}\right) \cos ^{2}\left(\frac{w}{2}\right) d w\)
View solution