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Chapter 12: Problem 2

Man bestimme das Integral $$ I=\int_{-\pi}^{\pi} \frac{\sinh x \cos x}{1+x^{2}} \mathrm{~d} x $$

Short Answer

Expert verified
Question: What is the value of the definite integral $$I=\int_{-\pi}^{\pi} \frac{\sinh x \cos x}{1+x^{2}} \mathrm{~d} x$$ Answer: The value of the integral is \(I = 0\).

Step by step solution

01

Check if the integrand is odd or even

First, let's analyze the integrand to see if it is an odd function or even function. A function \(f(x)\) is even if \(f(x) = f(-x)\). A function \(f(x)\) is odd if \(f(-x) = -f(x)\). In this case, we have: $$ f(x) = \frac{\sinh x \cos x}{1+x^{2}} $$ Now we will find the value of \(f(-x)\): $$ f(-x) = \frac{\sinh(-x) \cos(-x)}{1+(-x)^{2}} $$ Since \(\sinh(-x) = -\sinh x\) and \(\cos(-x) = \cos x\), we have: $$ f(-x) = \frac{-\sinh x \cos x}{1+x^{2}} $$ It can be seen that \(f(-x) = -f(x)\), hence the given function is an odd function.
02

Use the property of odd functions

Since the given function is odd, and we are asked to evaluate the definite integral over a symmetric interval around the origin \([-\pi, \pi]\), we can use the property of odd functions. The property states that the definite integral of an odd function over a symmetric interval around the origin is always 0: $$ \int_{-a}^{a} f(x) \, \mathrm{d} x = 0 $$
03

Apply the property and find the integral value

Now, we can apply the property of odd functions to the given integral: $$ \int_{-\pi}^{\pi} \frac{\sinh x \cos x}{1+x^{2}} \mathrm{~d} x = 0 $$ So, the integral value \(I\) is: $$ I = 0 $$

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Most popular questions from this chapter

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