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Magazine Chapter 7: Problem 15
Perform the indicated integrations. \(\int_{0}^{\pi / 4} \frac{\cos x}{1+\sin ^{2} x} d x\)
Short Answer
Expert verified
The answer is \( \frac{\pi}{8} \).
Step by step solution
01
Identify the Integration Technique
The integral \( \int_{0}^{\pi / 4} \frac{\cos x}{1+\sin ^{2} x} \ dx \) can be solved using substitution. Notice that the derivative of \( \sin x \) is \( \cos x \), suggesting substitution might simplify the process.
02
Choose a Substitution
Let \( u = \sin x \). Then, \( du = \cos x \, dx \). Substitute these into the integral, which changes the limits of integration. When \( x = 0 \), \( u = \sin 0 = 0 \) and when \( x = \frac{\pi}{4} \), \( u = \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2} \).
03
Rewrite the Integral
Substitute the chosen values into the integral. The integral becomes \( \int_{0}^{\sqrt{2}/2} \frac{1}{1+u^2} \ du \). This is a standard form integral for the inverse tangent function.
04
Integrate Using a Known Formula
The integral \( \int \frac{1}{1+u^2} \, du \) is equal to \( \arctan(u) + C \). Therefore, Evaluate \( \int_{0}^{\sqrt{2}/2} \frac{1}{1+u^2} \, du = \arctan\left(\frac{\sqrt{2}}{2}\right) - \arctan(0) \).
05
Evaluate the Definite Integral
Calculate \( \arctan\left(\frac{\sqrt{2}}{2}\right) \) and \( \arctan(0) \). \( \arctan(0) = 0 \) and \( \arctan\left(\frac{\sqrt{2}}{2}\right) \) is a specific angle value based on the inverse tangent function. Use a calculator or consult a table if necessary.
06
Calculate Final Value
With \( \arctan(0) = 0 \), the integral simplifies to \( \arctan\left(\frac{\sqrt{2}}{2}\right) \). Approximating \( \arctan\left(\frac{\sqrt{2}}{2}\right) \), if exact, yields \( \frac{\pi}{8} \). Therefore, the value of the whole integral is \( \frac{\pi}{8} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integration by Substitution
Integration by substitution is a powerful technique. It helps make complicated integrals easier to solve. Imagine substitution as changing the perspective. Enter the world of the integral.Here's how it works:
- Choose the right substitution. Look for parts of the integral that are derivatives of each other. Like in the given exercise, the derivative of \( \sin x \) is \( \cos x \). This makes \( u = \sin x \) a good substitution.
- Transform the integrand and limits. With \( du = \cos x \, dx \), you replace \( \cos x \, dx \) \( = \) \( du \) in the integral. This simplifies calculations.
- Adjust the limits of integration as well. When \( x = 0 \), \( u = 0 \), and when \( x = \frac{\pi}{4} \), \( u = \frac{\sqrt{2}}{2} \).
Definite Integrals
Definite integrals calculate the area under a curve within a specific interval. They're essential for understanding continuous functions and accumulation processes.Some key points about definite integrals are:
- The integral is evaluated over a specified range. In this case, from \( x = 0 \) to \( x = \frac{\pi}{4} \).
- Limits of integration define where this journey of calculation begins and ends. Convert these limits during substitution. The transformation leads you from \( x \)-limits to \( u \)-limits.
- Once you integrate, plug the new limits back into the antiderivative. This evaluates the complete expression. Much like closing a story with a satisfying conclusion. In the example, we solved the integral \( \int_{0}^{\sqrt{2}/2} \frac{1}{1+u^2} \, du \).
Inverse Trigonometric Functions
Inverse trigonometric functions are essential when dealing with specific standard integrals. These functions undo what regular trigonometric functions do.Here's how they work:
- The inverse tangent function, represented as \( \arctan \), is crucial when integrating forms like \( \frac{1}{1+u^2} \). It tells you at what angle a given tangent value occurs.
- They appear naturally in our integrals after substitution. In this case, after transforming the integral into \( \int \frac{1}{1+u^2} \, du \), it matches the \( \arctan(x) \) base form.
- The specific value \( \arctan\left(\frac{\sqrt{2}}{2}\right) \) was calculated to find the area exactly. Use tables or calculators if needed to find exact values.
