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Chapter 8: Problem 7

Evaluate the following integrals. $$\int_{0}^{\pi / 2} \frac{\sin \theta}{1+\cos ^{2} \theta} d \theta$$

Short Answer

Expert verified
Question: Evaluate the integral $$\int_{0}^{\pi /2} \frac{\sin \theta}{1+\cos ^{2} \theta} d\theta$$. Answer: $$-\frac{\pi}{4}$$

Step by step solution

01

Perform Substitution

Let's make a substitution to simplify the integral. Take $$u = \cos \theta.$$ Then, differentiate u with respect to 胃 to get $$\frac{du}{d\theta} = -\sin \theta.$$ Now, we can express $$d\theta$$ in terms of $$du$$: $$d\theta = -\frac{du}{\sin \theta}.$$ By plugging this into our integral and updating the limits, we get $$\int_{1}^{0} \frac{\sin \theta}{1+\cos ^{2} \theta} \cdot -\frac{1}{\sin \theta} du = \int_{1}^{0} \frac{1}{1+u^2} du.$$
02

Evaluate the Resulting Integral

The integral is now simplified as $$\int_{1}^{0} \frac{1}{1+u^2} du.$$ The antiderivative of $$\frac{1}{1+u^2}$$ is $$\arctan(u) + C$$, where C is the constant of integration. So, we have $$\int_{1}^{0} \frac{1}{1+u^2} du = \arctan(u)\Big|_{1}^{0}$$
03

Evaluate the Integral Over the Given Interval

Now, we evaluate the integral over the given interval: $$\arctan(u)\Big|_{1}^{0} = \arctan(0) - \arctan(1)$$ Since $$\arctan(0) = 0$$ and $$\arctan(1) = \frac{\pi}{4}$$, we find $$\arctan(u)\Big|_{1}^{0} = 0 - \frac{\pi}{4} = -\frac{\pi}{4}.$$ Therefore, the integral evaluates to $$-\frac{\pi}{4}$$.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Trigonometric Integrals
Trigonometric integrals are a class of integrals that involve trigonometric functions. Solving these integrals typically requires knowledge of trigonometric identities, integration techniques, and sometimes creative substitutions.

For instance, integrals involving functions like \( \frac{\text{sin} \theta}{1+\text{cos}^2 \theta} \) can seem daunting due to the non-standard denominator. To tackle this, we can utilize trigonometric identities, or as in our example, substitution to simplify the integral. Sometimes these integrals may result in inverse trigonometric functions as their antiderivatives, which is the case in our given exercise.
U-Substitution Method
The u-substitution method is a powerful technique for evaluating integrals, essentially serving as the reverse chain rule in calculus. It's particularly useful when dealing with functions that are products of two separate functions, one of which is the derivative of the other.

To apply u-substitution, identify a section of the integrand that can be set as 'u', and then find 'du', which is the derivative of 'u' with respect to x (or in our case, \( \theta \)). Replace these parts in the integral, and update the limits if dealing with a definite integral. This transforms the integral into a usually more friendly form, as we saw in our exercise where \( u = \text{cos} \theta \) simplified the evaluation.
Definite Integral Evaluation
Definite integral evaluation involves finding the exact value of the integral over a specified interval. After finding the indefinite integral (antiderivative), we use the Fundamental Theorem of Calculus to evaluate it at the upper and lower bounds of the interval, which are often updated after a substitution.

It's important to take note of the updated limits when applying u-substitution, as we did in our exercise, where the limits changed from \( 0 \) to \( \frac{\text{pi}}{2} \) for \( \theta \) to \( 0 \) to \( 1 \) for \( u \). We then calculate the difference between the antiderivative evaluated at these new limits to determine the integral's value.
Antiderivatives
Antiderivatives, also known as indefinite integrals, are functions that represent the area under the curve of the given function but without specific bounds. Finding an antiderivative means determining a function, F(x), whose derivative is equal to the integrand.

In our exercise, we were looking for the antiderivative of \( \frac{1}{1+u^2} \), which is \( \text{arctan}(u) + C \), where C is a constant. When evaluating a definite integral, we don't need to worry about the constant, as it cancels out when computing the difference between the upper and lower bounds.

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Most popular questions from this chapter

Work Let \(R\) be the region in the first quadrant bounded by the curve \(y=\sec ^{-1} x\) and the line \(y=\pi / 3 .\) Suppose a tank that is full of water has the shape of a solid of revolution obtained by revolving region \(R\) about the \(y\) -axis. How much work is required to pump all the water to the top of the tank? Assume \(x\) and \(y\) are in meters.

An integrand with trigonometric functions in the numerator and denominator can often be converted to a rational function using the substitution \(u=\tan (x / 2)\) or, equivalently, \(x=2 \tan ^{-1} u .\) The following relations are used in making this change of variables. $$A: d x=\frac{2}{1+u^{2}} d u \quad B: \sin x=\frac{2 u}{1+u^{2}} \quad C: \cos x=\frac{1-u^{2}}{1+u^{2}}$$ $$\text { Evaluate } \int \frac{d x}{1+\sin x+\cos x}$$.

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. a. It is possible for a computer algebra system to give the result \(\int \frac{d x}{x(x-1)}=\ln (x-1)-\ln x\) and a table of integrals to give the result \(\int \frac{d x}{x(x-1)}=\ln \left|\frac{x-1}{x}\right|+C\) b. A computer algebra system working in symbolic mode could give the result \(\int_{0}^{1} x^{8} d x=\frac{1}{9},\) and a computer algebra system working in approximate (numerical) mode could give the result \(\int_{0}^{1} x^{8} d x=0.11111111\).

Comparing the Midpoint and Trapezoid Rules Compare the errors in the Midpoint and Trapezoid Rules with \(n=4,8,16,\) and 32 subintervals when they are applied to the following integrals (with their exact values given). $$\int_{0}^{1}\left(8 x^{7}-7 x^{8}\right) d x=\frac{2}{9}$$

A powerful tool in solving problems in engineering and physics is the Laplace transform. Given a function \(f(t),\) the Laplace transform is a new function \(F(s)\) defined by $$F(s)=\int_{0}^{\infty} e^{-s t} f(t) d t$$ where we assume s is a positive real number. For example, to find the Laplace transform of \(f(t)=e^{-t},\) the following improper integral is evaluated using integration by parts: $$F(s)=\int_{0}^{\infty} e^{-s t} e^{-t} d t=\int_{0}^{\infty} e^{-(s+1) t} d t=\frac{1}{s+1}$$ Verify the following Laplace transforms, where a is a real number. $$f(t)=\sin a t \rightarrow F(s)=\frac{a}{s^{2}+a^{2}}$$
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