Question

Preliminary steps The following integrals require a preliminary step such as a change of variables before using the method of partial fractions. Evaluate these integrals. $$\int \frac{\sec t}{1+\sin t} d t$$

Short answer

Question: Evaluate the integral \(\int \frac{\sec t}{1+\sin t} dt\). Answer: \(\ln |1 + \sin t| + C\)

Step-by-step solution

Rewrite the integrand using the identity \(\sec t = 1/\cos t\)

First, rewrite the given integral in terms of \(\cos t\). Using the identity \(\sec t = \frac{1}{\cos t}\), we have: $$\int \frac{\sec t}{1+\sin t} dt = \int \frac{\frac{1}{\cos t}}{1 + \sin t}dt $$

Choose the change of variable and find the differential

Make the substitution \(u = 1 + \sin t\). This requires finding the differential. Differentiate \(u=1+\sin t\) with respect to \(t\): $$\frac{d u}{dt} = \frac{d(1+ \sin t)}{dt} = 0 + \cos t$$ Now, solve for \(dt\) by dividing both sides by \(\cos t\): $$dt = \frac{du}{\cos t}$$

Substitute the variables in the integral

Substitute \(u=1+\sin t\) and \(dt = \frac{du}{\cos t}\) into the integral: $$\int \frac{\frac{1}{\cos t}}{1 + \sin t} dt = \int \frac{1}{u} \frac{du}{\cos t}$$

Simplify the integral and solve

Simplify the integral and integrate with respect to \(u\): $$\int \frac{1}{u} \frac{du}{\cos t} = \int \frac{1}{u} du$$ Performing this integration yields: $$\ln |u| + C = \ln |1 + \sin t| + C$$

Write the final answer

Remember to provide the final answer in terms of \(t\), so substitute back the variable \(u = 1 + \sin t\) into our result: $$\ln |1 + \sin t| + C$$ Hence, the integral \(\int \frac{\sec t}{1+\sin t} dt\) is equal to \(\ln |1 + \sin t| + C\).

Key concepts

Integration by Substitution

One potent technique in integral calculus is integration by substitution. It's the reverse process of the chain rule in differentiation and is used for simplifying integrals. The basic idea is to change the variable of integration to simplify the integral into a form that is easier to evaluate.

Consider an integral expressed as \[\int f(g(x))g'(x)dx\] We can simplify this by substituting \( u = g(x) \) and then replacing \( dx \) with \( du/g'(x) \). The integral would then be rewritten as \[\int f(u)du\] which might be simpler to evaluate. For example, in the given exercise, the substitution \( u = 1 + \sin t \) was used to simplify the integral involving trigonometric identities into an integral in terms of \( u \).

The choice of substitution is crucial and often requires practice to identify the most effective substitution that simplifies the integral the most efficiently.

Partial Fractions

Partial fractions are a technique used to integrate rational functions, or fractions of polynomials. If we have a fraction where the degree of the numerator is less than the degree of the denominator, we can express it as a sum of simpler fractions—the so-called partial fractions. To integrate complex rational expressions, we often break them down into partial fractions that we can integrate individually.

This method is mainly applicable when we deal with integrals of rational functions after polynomial long division, if necessary. Although not demonstrated in the provided exercise, it's important to remember that sometimes before employing substitution, converting the integrand into partial fractions may be necessary if it involves a complex rational expression.

Trigonometric Identities

Trigonometry is replete with trigonometric identities—equations that relate the trigonometric functions to one another. These identities simplify and solve trigonometric expressions, and they are essential when working with integrals involving trig functions. For instance, the most commonly used identities include the Pythagorean identities, angle sum and difference identities, double and half-angle identities, among others.

In the solved exercise, the identity \( \sec t = 1/\cos t \) is used to rewrite the given integral in a more workable form. Knowledge of how to manipulate these identities allows us to convert complex trigonometric integrals into simpler algebraic forms that can then easily be integrated, as shown in the step-by-step solution. Students should familiarize themselves with various trigonometric identities as they greatly ease the process of integration.

Differential Calculus

Differential calculus plays a vital role in computing integrals, especially when using the substitution method. This field of mathematics focuses on the rate at which quantities change, known as the derivative. Calculating a derivative is the fundamental operation of differential calculus.

In integrals that require substitution, as in our exercise, knowledge of differential calculus is required to find the differential of the new variable. For example, finding \( \frac{du}{dt} \) when \( u = 1 + \sin t \) is a direct application of differential calculus. This derivative, \( \cos t \) in our case, is then used to transform the \( dt \) in the original integral into terms involving \( du \) that match our substitution. Regardless of the complexity of the function being integrated, understanding the relationship between derivatives and integrals is key to mastering the art of integration.