Problem 93 - Calculus Early Transcendentals

Question

\(A n\) 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. \(A: d x=\frac{2}{1+u^{2}} d u \quad B: \sin x=\frac{2 u}{1+u^{2}} \quad C: \cos x=\frac{1-u^{2}}{1+u^{2}}\) $$\text { Evaluate } \int_{0}^{\pi / 2} \frac{d \theta}{\cos \theta+\sin \theta}$$

Step-by-Step Solution

Answer

The value of the given integral is -1.

Step 1: Write down the given integral and substitution formulas

The given integral is: $$\int_{0}^{\pi / 2} \frac{d \theta}{\cos \theta+\sin \theta}$$ Substitution formulas provided are: $$dx=\frac{2}{1+u^{2}} du \quad (A)$$ $$\sin x=\frac{2 u}{1+u^{2}} \quad (B)$$ $$\cos x=\frac{1-u^{2}}{1+u^{2}} \quad (C)$$

Step 2: Use substitution \(u = \tan(\frac{x}{2})\) to transform the given integral

Substitute \(\sin \theta\) and \(\cos \theta\) by using formulas (B) and (C), and substitute \(d\theta\) by using formula (A), we have: $$\int \frac{\frac{2}{1+u^{2}} du}{\frac{1-u^2}{1+u^{2}}+\frac{2u}{1+u^{2}}}$$ Simplify the integrand: $$\int \frac{2}{(1-u^{2})+2u} du$$

Step 3: Integrate the rational function

To integrate the rational function, we can use partial fraction decomposition. First, we factor the denominator: $$2 - 4u + u^2 = (2-u)^2$$ Then, the partial fraction decomposition of the integrand is: $$\frac{2}{(2-u)^{2}} = A$$ Solve for \(A\), we have: $$A = 2$$ Now, we can integrate the partial fraction: $$\int \frac{2}{(2-u)^{2}} du = -\frac{2}{2-u} + C$$

Step 4: Substitute back for the original variable

We now substitute back \(x\) for \(u\). Recall that \(u = \tan(\frac{x}{2})\), so \(\frac{x}{2} = \tan^{-1}(u)\) and \(x = 2\tan^{-1}(u)\). Thus, the antiderivative is: $$-\frac{2}{2-\tan(\frac{x}{2})} + C$$

Step 5: Evaluate the definite integral

Now, we need to evaluate the definite integral from \(0\) to \(\frac{\pi}{2}\): $$\int_{0}^{\frac{\pi}{2}} \frac{d \theta}{\cos \theta+\sin \theta} = \left[-\frac{2}{2-\tan(\frac{x}{2})}\right]_{0}^{\frac{\pi}{2}}$$ Evaluate the antiderivative at the bounds: $$\left[-\frac{2}{2-\tan(\frac{\pi}{2}{2})}\right] - \left[-\frac{2}{2-\tan(\frac{0}{2})}\right] = -\frac{2}{1} - (-\frac{2}{2}) = -2 + 1 = -1$$ The value of the given integral is \(-1\).

Key Concepts

Substitution Method

When dealing with trigonometric integrals, especially those involving both sine and cosine, the substitution method can be quite helpful. This technique often involves introducing a new variable to simplify the integration process.

In the context of our original exercise, the substitution used is \( u = \tan(\frac{x}{2}) \). This specific substitution is particularly useful when the integral involves both sine and cosine. It leverages the identities:

• \( \sin x = \frac{2u}{1+u^2} \)
• \( \cos x = \frac{1-u^2}{1+u^2} \)

These identities allow us to transform the original trigonometric integral into a rational format that's easier to handle. This method not only simplifies the trigonometric components but also transforms the integral into a form where other techniques, like polynomial integration or partial fraction decomposition, can be applied.

Using substitution might seem complex at first, but with practice, you'll find it to be a powerful tool in tackling intricate integrals.

Partial Fraction Decomposition

Once a trigonometric integral is converted into a rational function using substitution, partial fraction decomposition becomes a valuable tool in the integration process. This technique involves expressing the rational function as a sum of simpler fractions that are easier to integrate.

For example, after our substitution in the original problem, we obtained an integral of the form \( \frac{2}{(1-u^2) + 2u} \). Further simplification transformed the denominator into \((2-u)^2\), allowing us to use partial fraction decomposition.

The process involves:

• Factoring the denominator, if possible.
• Expressing the integrand as a sum of fractions with simpler denominators.
• Solving for the coefficients of these partial fractions.

In the context of our exercise, the partial fraction decomposition was straightforward as the integrand simplified to a single simple fraction \( \frac{2}{(2-u)^2} \). After determining the coefficients, the integration becomes a matter of applying basic antiderivative formulas.

Mastering partial fraction decomposition can greatly expand your mathematical toolkit, especially when dealing with challenging integrals.

Change of Variables

After integrating using the methods above, it is crucial to revert back to the original variable of the integral. This process is known as the change of variables. In substitution, you've transformed the variables to simplify the math, but the final result needs to reflect the original terms and conditions of the integral.

In the exercise at hand, after having successfully integrated with respect to \( u \), we have: \(-\frac{2}{2-u}\). But remember, \( u \) was introduced by \( u = \tan(\frac{x}{2}) \), which means we must now express the solution in terms of \( x \). This is often done by substituting back the relation \( x = 2 \tan^{-1}(u) \) to find the antiderivative in terms of \( x \).

Ultimately, in definite integrals such as this, once you substitute back, evaluate the limits within the transformed function. This gives the final answer to the problem.

Understanding and effectively applying change of variables ensures that your solution addresses the question as it was originally presented and adheres to any given bounds or conditions.