Question

In the following exercises, evaluate the definite integral. $$ \int_{0}^{\pi / 3} \frac{\sin x-\cos x}{\sin x+\cos x} d x $$

Short answer

The integral evaluates to \(-\ln\left(\frac{\sqrt{3} + 1}{2}\right)\)."

Step-by-step solution

Identify the Integral

The integral that we need to evaluate is \( \int_{0}^{\pi / 3} \frac{\sin x - \cos x}{\sin x + \cos x} \; dx \). It is a definite integral, so we will evaluate the integral within the limits from 0 to \( \frac{\pi}{3} \).

Simplify the Integrand

To evaluate the integral, simplify the integrand \( \frac{\sin x - \cos x}{\sin x + \cos x} \). We can use a trigonometric identity or substitution to make it simpler. Consider the substitution \( u = \sin x + \cos x \), which means \( du = (\cos x - \sin x) \, dx \). This simplifies the integrand.

Change the Variable

Using the substitution \( u = \sin x + \cos x \), we can express the integral in terms of \( u \). The differential term becomes \( du = (\cos x - \sin x) \, dx \), and hence \( dx = \frac{du}{\cos x - \sin x} \). Substitute these into the integral, transforming it to \( \int \frac{-1}{u} \, du \).

Integrate with Respect to the New Variable

Integrating \( \int \frac{-1}{u} \, du \) gives the result \( -\ln |u| + C \), where \( C \) is the constant of integration. Since this is a definite integral, we apply the limits afterwards.

Change Back to the Original Variable

Substitute back \( u = \sin x + \cos x \) into the integrated result to get \( -\ln|\sin x + \cos x| \).

Evaluate the Definite Integral

Use the limits to evaluate the definite integral: \( \int_{0}^{\pi/3} \frac{\sin x - \cos x}{\sin x + \cos x} \, dx = [-\ln|\sin x + \cos x|]_{0}^{\pi/3} \). Calculate this as: - At \( x = \pi/3 \), \( \sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2} \) and \( \cos(\frac{\pi}{3}) = \frac{1}{2} \), so value becomes \( -\ln|\frac{\sqrt{3}}{2} + \frac{1}{2}| \).- At \( x = 0 \), \( \sin(0) = 0 \) and \( \cos(0) = 1 \), thus value is \( -\ln|1| \). Calculate these values and subtract.

Calculate and Finalize Result

The expression simplifies as:\[-\ln\left(\frac{\sqrt{3} + 1}{2}\right) + \ln(1) = -\ln\left(\frac{\sqrt{3} + 1}{2}\right)\].Since \( \ln(1) = 0 \), the definite integral evaluates to \( -\ln\left(\frac{\sqrt{3} + 1}{2}\right) \).

Key concepts

Trigonometric Substitution

Trigonometric substitution is a powerful technique when evaluating integrals involving trigonometric expressions. In this exercise, we simplify the integrand \( \frac{\sin x - \cos x}{\sin x + \cos x} \) by applying a clever substitution. By letting \( u = \sin x + \cos x \), we transform the expression to one that is easier to integrate. This substitution not only simplifies the integrand but also allows us to express the differential \( dx \) in terms of \( u \), making the integral manageable.
Here's how it works:

• Substitute \( u = \sin x + \cos x \)
• Find the differential \( du = (\cos x - \sin x) \, dx \)
• Express \( dx = \frac{du}{\cos x - \sin x} \)
• Simplify the integral to \( \int \frac{-1}{u} \, du \)

By choosing the right trigonometric substitution, we turn a complex problem into a simple logarithmic integration.

Limits of Integration

Limits of integration define the boundary values of variables for definite integrals. They are crucial components that ensure we evaluate the integral over the correct interval. In this exercise, we deal with the definite integral \( \int_{0}^{\pi / 3} \frac{\sin x - \cos x}{\sin x + \cos x} \, dx \). The limits of integration are 0 and \( \frac{\pi}{3} \).
Understanding the limits:

• The lower limit 0 represents the starting point for our integral.
• The upper limit \( \frac{\pi}{3} \) indicates where integration stops.
• After substitution, it’s important to evaluate these limits in terms of the new variable \( u \).
• Once we return to the original variable, we apply these limits to compute the definite integral's value.

The limits ensure that all transformations and simplifications respect the original interval, ensuring an accurate result.

Logarithmic Integration

Logarithmic integration arises often when the integrand can be represented as a fraction. It becomes particularly straightforward when the derivative of the denominator appears in the numerator. In this exercise, after applying trigonometric substitution, we arrive at the integral \( \int \frac{-1}{u} \, du \), which is a perfect candidate for logarithmic integration.
Here's the simple process:

• Recognize the form \( \int \frac{-1}{u} \, du \) as the natural logarithm of \( u \).
• The antiderivative is \( -\ln |u| + C \).
• For definite integrals, you apply the limits of integration directly.
• Convert back to the original variable to express the solution in terms of \( x \).

Logarithmic integration is efficient and effective here, converting our problem into a straightforward application of a logarithmic function, leading to the final result.