Question

Evaluate the following integrals. Include absolute values only when needed. $$\int_{0}^{\pi / 2} \frac{\sin x}{1+\cos x} d x$$

Short answer

Question: Evaluate the definite integral \(\int_{0}^{\pi / 2} \frac{\sin x}{1+\cos x} d x\). Answer: \(\int_{0}^{\pi / 2} \frac{\sin x}{1+\cos x} d x = \ln{2}\).

Step-by-step solution

Simplify the integrand if possible

We take a closer look at the integrand \(\frac{\sin x}{1+\cos x}\). There is no apparent way to simplify this expression, so we'll proceed with integration techniques.

Find the antiderivative

We're trying to find the antiderivative of the given function, which requires some creative substitution. Let's substitute \(u = 1 + \cos x\) and see how it goes. $$ u = 1 + \cos x \Rightarrow du = \frac{du}{dx} = -\sin x \, dx $$ Now we can rewrite the integral in terms of \(u\): $$ \int_{0}^{\pi / 2} \frac{\sin x}{1+\cos x} d x = -\int \frac{1}{u} du $$ The antiderivative of \(\frac{1}{u}\) is \(\ln |u|\), so the expression becomes: $$ -\ln |u| + C = -\ln |1+\cos x| + C $$

Evaluate the definite integral

Now, we need to evaluate the definite integral within the limits \(0\) and \(\pi/2\): $$ \int_{0}^{\pi / 2} \frac{\sin x}{1+\cos x} d x = -[\ln |1+\cos(\pi/2)| - \ln |1+\cos(0)|] $$ By evaluating the cosine values in the expression, we get: $$ -\ln|(1+0)| + \ln|(1+1)| $$ This simplifies to: $$ -\ln{1} + \ln{2} = 0 + \ln{2} = \ln{2} $$ So the definite integral evaluates to: $$ \int_{0}^{\pi / 2} \frac{\sin x}{1+\cos x} d x = \ln{2} $$ Since the answer is a logarithmic function, there's no need to include absolute values in our solution.

Key concepts

Trigonometric Substitution

Trigonometric substitution is a powerful technique used to simplify integrals involving trigonometric functions. It often involves replacing parts of an integral with a trigonometric function that is easier to handle.
For example, in this exercise, we did not directly use trigonometric substitution as our primary approach to integration, but we introduced the concept through the substitution \( u = 1 + \cos x \).
This often means spotting relationships such as \( \sin^2(x) + \cos^2(x) = 1 \) and expressing elements of a problem using these identities.
Using basic trigonometric identities like \( 1 + \cos x \) helps us manipulate the integral more easily by transforming it into a different variable that is less complex to work with.
Though it can sometimes make the integral look more complicated at first glance, it usually serves to simplify the problem when the right substitutions are used.

• Identify parts of the function that match trigonometric identities.
• Replace those parts with an appropriate substitution that reduces complexity.
• Solve the resulting integral using the new variable.

Natural Logarithm

The natural logarithm, represented by \( \ln \), is a logarithm to the base \( e \), where \( e \) is the mathematical constant approximately equal to 2.71828. In calculus, the natural logarithm appears frequently in integration, especially in cases like this one where the integral takes the form \( \int \frac{1}{u} \, du \).
In the given exercise, after making a substitution, the integral turned into \( \int \frac{1}{u} \, du \), whose antiderivative is \( \ln |u| + C \). Here, \( C \) is the constant of integration, usually omitted in definite integrals.
Using the natural logarithm helps simplify the integration process, especially for rational functions like \( \frac{1}{u} \). It often provides a neat way to represent the area under a curve when an antiderivative cannot be expressed as elementary functions.

• The natural logarithm \( \ln \) is essential when integrating functions of the form \( \frac{1}{u} \).
• It simplifies the representation of the antiderivative.
• It commonly appears when working with logarithmic and exponential decay problems.

Integration by Substitution

Integration by substitution is a method used to find integrals by changing the variable of integration to simplify the integral. This technique is the reverse process of the chain rule used in differentiation.
In our exercise, we used substitution by letting \( u = 1 + \cos x \), which simplifies the original integral \( \int_{0}^{\pi/2} \frac{\sin x}{1+\cos x} \, dx \) into a more straightforward form \( -\int \frac{1}{u} \, du \).
This substitution is strategic because it transforms the integral into a basic form that is easier to integrate and handle. After finding the antiderivative, we then substitute back the original variables and evaluate the definite integral over the specified interval.
To effectively use integration by substitution, you should:

• Identify a substitution \( u \) that makes the integral easier to solve.
• Compute \( du \) and replace \( dx \) or any other differential accordingly.
• Integrate with respect to \( u \) and then transform back to the original variable if needed.
• Evaluate the resulting antiderivative within the integration limits if it is a definite integral.

Integration by substitution requires some creativity in choosing an appropriate substitution, but with practice, it becomes an invaluable tool in solving a variety of calculus problems.