Problem 89 - 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. \(A: d x=\frac{2}{1+u^{2}} d u \quad B: \sin x=\frac{2 u}{1+u^{2}} \quad C: \cos x=\frac{1-u^{2}}{1+u^{2}}\) $$\text { Evaluate } \int \frac{d x}{1+\sin x}$$

Step-by-Step Solution

Answer

The substitution made in order to evaluate this integral is \(u = \tan(x/2)\) or \(x = 2 \tan^{-1} u\).

Step 1: Apply the substitution and rewrite the integral in terms of \(u\)

Applying the substitution \(u = \tan(x/2)\), we can rewrite the integrand using the given relations \(B\) and \(A\). \(\sin x = \frac{2u}{1+u^2}\) \(dx = \frac{2}{1+u^2} du\) So, the integral becomes: $$\int \frac{dx}{1 + \sin x} = \int \frac{\frac{2}{1+u^2} du}{1 + \frac{2u}{1+u^2}}$$

Step 2: Simplify the integral into a rational function

Next, let's simplify the integral: $$\int \frac{\frac{2}{1+u^2} du}{1 + \frac{2u}{1+u^2}} = \int \frac{2}{(1+u^2) + 2u} du$$

Step 3: Perform the integration

Now, let's integrate the rational function: $$\int \frac{2}{(1+u^2) + 2u} du = \int \frac{2}{u^2 + 2u + 1} du$$ We can rewrite the denominator as a square: $$ \int \frac{2}{(u+1)^2} du$$ Let's perform a substitution: \(v = u + 1\), so \(dv = du\): $$ \int \frac{2}{v^2} dv$$ Now, we can easily integrate this function: $$ 2 \int \frac{1}{v^2} dv = -\frac{2}{v} + C$$

Step 4: Substitute back to find the answer in terms of \(x\)

Substitute our \(v\) back to \(u+1\): $$-\frac{2}{u+1} + C$$ Finally, substitute \(u = \tan(x/2)\): $$-\frac{2}{\tan(x/2)+1} + C = -\frac{2}{(\tan(x/2)+1)} + C$$ So the final answer is: $$-\frac{2}{(\tan(x/2)+1)} + C$$

Key Concepts

Trigonometric Substitution

Trigonometric substitution is a valuable technique used in integration, especially when dealing with integrands that involve trigonometric functions. In particular, when you see an integral involving trigonometric expressions in the numerator or the denominator, invoking a substitution involving tangent can simplify the problem significantly.

When using trigonometric substitution, you replace a trigonometric expression in the integrand with a rational function. This approach often leads to a more manageable form for integration. One common substitution is based on setting \( u = \tan(x/2) \), which assists in converting sine and cosine expressions into rational ones. Here are the key identities used in such transformations:

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

These identities help change the form of the integral into something more treatable, allowing the use of algebraic manipulation and standard integration techniques. This transformation is particularly useful because it turns complex trigonometric integrals into rational functions that are often easier to integrate directly.

Rational Functions

Rational functions are expressions of the form \( \frac{p(x)}{q(x)} \), where \( p(x) \) and \( q(x) \) are polynomials. When integrals involve trigonometric functions that can be converted into rational functions, It can greatly simplify the integration process.

In the step-by-step solution, the integral \( \int \frac{dx}{1 + \sin x} \) is transformed into a rational function of \( u \) through substitution. After applying \( u=\tan(x/2) \), the integrand becomes \( \int \frac{2}{u^2 + 2u + 1} du \).

The key here is to identify expressions within the integrand that can be rewritten as simpler polynomials. Upon simplification, we found that \( u^2 + 2u + 1 \) is a perfect square, specifically \( (u+1)^2 \). This realization made the function easier to integrate, illustrating why converting to a rational function through substitutions could be powerful. It allows us to focus on polynomial algebra, which is typically more straightforward than trigonometric calculus.

Definite and Indefinite Integrals

Integrals are a fundamental concept in calculus, representing the area under a curve. They can be definite, having limits of integration, or indefinite, lacking such limits and yielding a general antiderivative.

Indefinite integrals, like the one solved in the exercise \( \int \frac{2}{(u+1)^2} du \), focus on finding a function whose derivative gives the integrand. The solution process often involves substitution and identification of patterns that simplify into known integrals. In this context, substituting back the original variable converts our result from the integral of a transformed function to an expression in terms of the initial variable \( x \).

For indefinite integrals, the solution includes an arbitrary constant \( C \) since many functions can differ by a constant yet share the same derivative. This constant accounts for all possible vertical shifts of the antiderivative.

Understanding definite integrals involves calculating actual values by evaluating the integral from one limit to another, focusing less on patterns and more on numerical accuracy. The techniques of substitution highlighted here transfer across both types, aiding in breaking down complex integrals across different scenarios.