Question

The integral \(\int_{0}^{2 \pi} \frac{d x}{5-3 \cos x}\) is readily taken with the aid of the substitution \(\tan \frac{x}{2}=z\). We have \(\int_{0}^{2 \pi} \frac{d x}{5-3 \cos x}=\int_{0}^{0} \frac{2 \mathrm{dz}}{\left(1+z^{2}\right)\left(5-3 \frac{1-z^{2}}{1+z^{2}}\right)}=0\)

Short answer

Short Answer: By applying the given substitution \(\tan{\frac{x}{2}}=z\) and evaluating the integral with respect to z, we find that the integral is \(\int_{0}^{0} \frac{2 dz}{\left(1+z^{2}\right)\left(5-3 \frac{1-z^{2}}{1+z^{2}}\right)}\), which evaluates to zero since the limits of integration are the same (0 to 0). This confirms that the integral of the given function is equal to zero using the substitution method.

Step-by-step solution

Substitution

Start by substituting \(\tan{\frac{x}{2}}=z\) as given. To change the limits of integration, we need to find \(x\) in terms of \(z\) as well as its differential. We can use the following identities: 1. \(x = 2\arctan{z}\) 2. \(\cos{x} = \frac{1 - z^2}{1 + z^2}\) Now, let's find \(\frac{dx}{dz}\) using the relation from point 1: \(\frac{dx}{dz} = \frac{d}{dz}\left(2\arctan{z}\right) = \frac{2}{1 + z^2}\) Now we can evaluate the new limits of integration: \(x=0 \Rightarrow z = \tan{0} = 0\) \(x=2\pi \Rightarrow z = \tan{\pi} = 0\) The integral in terms of \(z\) is: $$\int_{0}^{0} \frac{2 dz}{\left(1+z^{2}\right)\left(5-3 \frac{1-z^{2}}{1+z^{2}}\right)}$$

Integrating in terms of z

Evaluate the integral: $$\int_{0}^{0} \frac{2 dz}{\left(1+z^{2}\right)\left(5-3 \frac{1-z^{2}}{1+z^{2}}\right)} = 0$$ The integral evaluates to zero because the upper and lower limits of integration are the same, which means the integral has no width. This confirms that the integral of the given function \(\int_{0}^{2 \pi} \frac{d x}{5-3 \cos x}\) is indeed equal to zero when using the substitution \(\tan \frac{x}{2}=z\).

Key concepts

Trigonometric Substitution

Trigonometric substitution is a useful integration technique, especially when dealing with integrals that contain square roots or certain quadratic expressions. The trigonometric substitution process involves replacing an algebraic expression with a trigonometric function to simplify the integral. It relies on trigonometric identities to change the form of the integral, making it easier to evaluate.

For example, in the exercise \[ \int_{0}^{2 \pi} \frac{dx}{5-3 \cos x} \], we use the substitution \( \tan \frac{x}{2}=z \) to transform the integral into a form that can be integrated using standard techniques. Here, we take advantage of the identity \( \cos x = \frac{1-z^2}{1+z^2} \) to rewrite the integrand in terms of \(z\), which simplifies the problem significantly.

Trigonometric substitution is particularly useful for problems where the integrand includes terms like \(\sqrt{a^2 - x^2}\), \(\sqrt{a^2 + x^2}\), or \(\sqrt{x^2 - a^2}\), as these lend themselves well to substitution with sine, tangent, or secant functions respectively.

Definite Integrals

Definite integrals calculate the signed area under the curve of a function, between two given points. The fundamental theorem of calculus connects differentiation and integration, stating that if we have a continuous function \(f(x)\) on an interval \[a, b\], then the definite integral of \(f(x)\) with respect to \(x\) from \(a\) to \(b\) can be found by evaluating the antiderivative of \(f(x)\) at \(b\) and \(a\) and then calculating the difference.

In the provided exercise, we see a definite integral with the same lower and upper limits of integration, which means we are looking at the area under a curve between identical points, essentially a 'zero-width' path. Therefore, the integral \[ \int_{0}^{0} \frac{2dz}{\left(1+z^{2}\right)\left(5-3 \frac{1-z^{2}}{1+z^{2}}\right)} \] simplifies to zero. This illustrates an important property of definite integrals: if the limits of integration are equal, the value of the integral is always zero.

Integration Techniques

Various integration techniques are essential for solving complex integrals that are not straightforward. This includes methods like substitution, integration by parts, partial fractions, and trigonometric substitution as presented in our example. Each method has its own set of scenarios under which it is most useful, and selecting the appropriate technique can simplify the process dramatically.

In our exercise, we used the method of trigonometric substitution, a choice that often comes in handy when the integrand involves trigonometric functions. Substitution, in general, is a versatile technique that applies well to a broad range of problems beyond just trigonometric integrals. It involves changing the variable of integration and sometimes the limits as well, in order to convert the integral into a simpler form that can often be evaluated using standard antiderivatives.

Limits of Integration

The limits of integration define the interval over which a definite integral is evaluated and are crucial in determining the result. Upon making a variable substitution in an integral, it's necessary to convert the original limits to correspond to the new variable. This step ensures that we're evaluating the integral over the appropriate range corresponding to the original problem.

In the given step-by-step solution, after substituting \(z = \tan \frac{x}{2}\), we find the new limits of integration by solving \(x = 0\) and \(x = 2\pi\) for \(z\). Both yield \(z = 0\), so the limits of integration for \(z\) are from 0 to 0. Typically, the limits would be different and the integration process would proceed normally; however, in this special case, the equal limits signify that the definite integral's value is zero, as there's no actual interval to calculate an area for.