Question

The values of the following integrals are known and can be found in integral tables or by computer. Your goal in evaluating them is to learn about contour integration by applying the methods discussed in the examples above. Then check your answers by computer. $${\int }_0^{\pi }\frac{d\theta }{(2+\cos \theta {)}^2}$$

Short answer

The value of the integral is $\frac{\pi }{9}$

Step-by-step solution

- Identify the Integral

The given integral is ${\int }_0^{\pi }\frac{d\theta }{(2+\cos \theta {)}^2}$.

- Use a Trigonometric Identity

Use the Weierstrass substitution $\tan (\frac{\theta }{2})=t$ which transforms the integral for trigonometric functions. Then $\cos \theta =\frac{1-t^2}{1+t^2}$ and $d\theta =\frac{2}{1+t^2}dt$.

- Transform the Integral

After substitution, the integral becomes: ${\int }_0^{\mathrm{\infty }}\frac{\frac{2}{1+t^2}dt}{{(2+\frac{1-t^2}{1+t^2})}^2}$. This simplifies to ${\int }_0^{\mathrm{\infty }}\frac{2dt}{(2(1+t^2)+(1-t^2){)}^2}$.

- Simplify the Expression

Simplify the denominator: $2+\cos \theta =2+\frac{1-t^2}{1+t^2}=\frac{2(1+t^2)+(1-t^2)}{1+t^2}=\frac{3+3t^2}{1+t^2}$. Thus, the integral becomes ${\int }_0^{\mathrm{\infty }}\frac{2dt}{(\frac{3(1+t^2)}{1+t^2}{)}^2}={\int }_0^{\mathrm{\infty }}\frac{2dt}{9(1+t^2)}$.

- Evaluate the Simplified Integral

The integral simplifies further: ${\int }_0^{\mathrm{\infty }}\frac{2dt}{9(1+t^2)}$. Factor out the constant term: $\frac{2}{9}{\int }_0^{\mathrm{\infty }}\frac{dt}{1+t^2}$. This is a standard integral which evaluates to $\frac{2}{9}{[{\tan }^{-1}(t)]}_0^{\mathrm{\infty }}$.

- Apply Limits

Substitute the bounds: $\frac{2}{9}[{\tan }^{-1}(\mathrm{\infty })-{\tan }^{-1}(0)]=\frac{2}{9}[\frac{\pi }{2}-0]=\frac{2}{9}\cdot \frac{\pi }{2}=\frac{\pi }{9}$.

Key concepts

trigonometric substitution

Trigonometric substitution is a technique used in calculus to simplify integrals involving square roots and trigonometric functions. By substituting a trigonometric function for a variable, the integrand often becomes easier to integrate. In this exercise, we use the Weierstrass substitution, where we set $\tan (\frac{\theta }{2})=t$. This substitution transforms the integral involving trigonometric functions into a rational function, making it simpler to handle. For example, $\cos \theta =\frac{1-t^2}{1+t^2}$ and $d\theta =\frac{2dt}{1+t^2}$. These transformations help convert the original integral ${\int }_0^{\pi }\frac{d\theta }{(2+\cos \theta {)}^2}$ into an easier form.

integral tables

Integral tables are an invaluable resource for anyone studying calculus. They provide a comprehensive list of integrals and their respective solutions, saving time and effort. You can find standard integrals, which might be challenging to solve by hand, in integral tables. In our step-by-step solution, we use the integral table for the integral ${\int }_0^{\mathrm{\infty }}\frac{dt}{1+t^2}$, which is a well-known integral that evaluates to $\frac{\pi }{2}$. Once we simplify our original integral to this standard form, the value can be looked up directly in the integral tables. This makes solving integrals faster and easier, especially for complex functions.

simplification of integrals

Simplifying integrals is a critical step in making complex integrals manageable. It involves algebraic manipulation and substitution to transform the integrand into a simpler form. In our exercise, after substituting $\tan (\frac{\theta }{2})=t$, we simplify the denominator of the integrand. The original integral ${\int }_0^{\pi }\frac{d\theta }{(2+\cos \theta {)}^2}$ was transformed to ${\int }_0^{\mathrm{\infty }}\frac{2dt}{9(1+t^2)}$. This simplification is crucial as it allows us to then factor out the constants and use known standard integrals from integral tables. The simplified integral becomes $\frac{2}{9}{\int }_0^{\mathrm{\infty }}\frac{dt}{1+t^2}$, which is much easier to evaluate.

Weierstrass substitution

The Weierstrass substitution, also known as the tangent half-angle substitution, is a powerful technique in integral calculus. It's particularly useful for integrals involving trigonometric functions. By setting $\tan (\frac{\theta }{2})=t$, you transform trigonometric functions into rational functions of $t$. For instance, under Weierstrass substitution, $\cos \theta$ becomes $\frac{1-t^2}{1+t^2}$, and $d\theta$ becomes $\frac{2dt}{1+t^2}$. These transformations help convert the integral into a form that's simpler to integrate. Using this substitution in our exercise allowed us to transform ${\int }_0^{\pi }\frac{d\theta }{(2+\cos \theta {)}^2}$ into ${\int }_0^{\mathrm{\infty }}\frac{2dt}{9(1+t^2)}$, making the integral more straightforward to evaluate.