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\(A n\) integrand with trigonometric functions in the numerator and denominator can often be converted to a rational integrand 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. \(A: d x=\frac{2}{1+u^{2}} d u \quad B: \sin x=\frac{2 u}{1+u^{2}} \quad C: \cos x=\frac{1-u^{2}}{1+u^{2}}\) $$\text { Evaluate } \int_{0}^{\pi / 3} \frac{\sin \theta}{1-\sin \theta} d \theta$$

Short Answer

Expert verified
Question: Evaluate the following integral: $$\int_{0}^{\pi / 3} \frac{\sin \theta}{1-\sin \theta} d \theta$$ Answer: $$2[\ln(\frac{\sqrt{3}+1}{\sqrt{3}-1})]$$

Step by step solution

01

Change the Variable

Using the substitution \(u = \tan(\frac{x}{2})\), we will rewrite the integral in terms of \(u\). First, let's find the new limits of integration. When \(x = 0\), we have \(u = \tan(\frac{0}{2}) = 0\). When \(x = \frac{\pi}{3}\), we have \(u = \tan(\frac{\pi}{3}{2}) = \tan(\frac{\pi}{6}) = \frac{1}{\sqrt{3}}\). Now we use the relations A, B, and C provided to rewrite the integrand. From relation B, we have \(\sin x = \frac{2u}{1 + u^2}\). So, \(1 - \sin x = 1 - \frac{2u}{1 + u^2} = \frac{1 - u^2}{1 + u^2}\). From relation A, we know that \(dx = \frac{2}{1 + u^2}du\). So the new integral becomes: $$\int_{0}^{\frac{1}{\sqrt{3}}} \frac{\frac{2u}{1+u^{2}}}{\frac{1-u^{2}}{1+u^{2}}} * \frac{2}{1+u^{2}} du$$
02

Simplify the New Integral

Simplifying the integral, we get: $$\int_{0}^{\frac{1}{\sqrt{3}}} \frac{2u}{1-u^{2}} * 2 du$$ $$= 4 \int_{0}^{\frac{1}{\sqrt{3}}} \frac{u}{1-u^{2}} du$$
03

Integrate with respect to \(u\)

Now integrate the rational function with respect to \(u\). This can be done using partial fractions. $$4 \int_{0}^{\frac{1}{\sqrt{3}}} \frac{u}{1-u^{2}} du = 4 \int_{0}^{\frac{1}{\sqrt{3}}} \frac{1}{2}(\frac{1}{1 - u} + \frac{1}{1 + u}) du$$ Now, integrate each term: $$= 4 (\frac{1}{2} [\int_{0}^{\frac{1}{\sqrt{3}}} \frac{1}{1 - u} du + \int_{0}^{\frac{1}{\sqrt{3}}} \frac{1}{1 + u} du])$$ $$= 2[-\ln |1 - u| + \ln |1 + u| ] \Big|_0^{\frac{1}{\sqrt{3}}}$$
04

Evaluate the Definite Integral

Now we simply need to evaluate the integral at the limits of integration: $$= 2[-\ln\left |\frac{1}{\sqrt{3}} - 1\right| + \ln\left |\frac{1}{\sqrt{3}} + 1\right | - (-\ln |1 - 0| + \ln |1 + 0|)]$$ $$= 2[\ln\left(\frac{\sqrt{3}+1}{\sqrt{3}-1}\right)]$$ So, the result of the integral is: $$\int_{0}^{\pi / 3} \frac{\sin \theta}{1-\sin \theta} d \theta = 2[\ln(\frac{\sqrt{3}+1}{\sqrt{3}-1})]$$

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Key Concepts

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

Integrals of Trigonometric Functions
Integrals involving trigonometric functions are a staple in calculus.

When dealing with integrals like \( \int \sin(x)dx \) or \( \int \cos(x)dx \), the process is straightforward. However, integrals where trigonometric functions are mixed with other types of functions, such as rational functions, can be more challenging. For these trickier cases, clever substitution methods are often used.

One such method involves using trigonometric identities to transform the integrand into a more manageable form. For instance, using the half-angle identity \( u = \tan(\frac{x}{2}) \) can be extremely helpful when the integrand includes a combination of \( \sin(x) \) and \( \cos(x) \) in a fraction. With this substitution, the challenging trigonometric integral turns into a rational function of \( u \), which is often much easier to integrate.

This method not only simplifies the integrand but also allows for the application of integration techniques like partial fractions, which are easier to handle with rational functions. Once the integral is computed in terms of \( u \), we just need to convert back to the original variable \( x \) to complete the problem.
Definite Integral Evaluation
Evaluating definite integrals involves computing the antiderivative of the integrand at the given interval and then calculating the difference of the antiderivative at the upper and lower limits.

To evaluate a definite integral of a function, one might need to apply various calculus techniques, especially when the function isn't straightforward. In the case of trigonometric functions, after simplifying the integral using appropriate substitutions and identities, the next step is to integrate and then plug in the limits.

When evaluating at the new limits post-substitution, it's crucial to ensure accuracy. This involves careful sign management, especially when dealing with trigonometric substitutions, since trigonometric functions can be positive or negative depending on the interval considered. Furthermore, it's important to remember to apply absolute value signs when necessary, particularly when taking the natural logarithm of a positive or negative quantity.

After the integration, the final step is to revert to the original variable if a substitution was made. This may involve additional steps, such as reapplying the inverse trigonometric function used in the substitution to revert to the original integral's limits.
Change of Variables in Integrals
Changing the variable in an integral, a technique also known as substitution, is a powerful tool to evaluate integrals that are initially difficult to handle.

The change of variables formula is given by \( \int f(g(x))g'(x)dx = \int f(u)du \) where \( u = g(x) \) and \( du = g'(x)dx \). This technique is particularly useful when direct integration is not feasible due to the complexity of the function. When changing variables, it is essential to also adjust the limits of integration if you're evaluating a definite integral.

Substitution involves finding a new variable \( u \) that simplifies the original integral. The choice of \( u \) should result in an integral that is easier to evaluate, often utilizing known integration methods such as integration by parts, partial fractions, or simple antiderivatives. After integrating with respect to \( u \), we then replace \( u \) with the original expression it stands for, completing the change of variables.

The effectiveness of this strategy lies in the fact that it can transform a complex problem into a simpler one, allowing for a solution that might otherwise require much more advanced techniques or even be unsolvable with elementary methods.

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Most popular questions from this chapter

Consider the solution of the logistic equation in Example 6. a. From the general solution \(\ln \left|\frac{P}{300-P}\right|=0.1 t+C,\) show that the initial condition \(P(0)=50\) implies that \(C=-\ln 5\). b. Solve for \(P\) and show that \(P=\frac{300}{1+5 e^{-0.1 t}}\).

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