Question

In the following exercises, evaluate the definite integral. $$ \int_{\pi / 6}^{\pi / 2} \csc x d x $$

Short answer

The definite integral evaluates to \(-\ln|2 + \sqrt{3}|\).

Step-by-step solution

Understand the Function

The given integral is \( \int_{\pi / 6}^{\pi / 2} \csc x \ dx \). The function \( \csc x \) is the cosecant function, which is the reciprocal of sine: \( \csc x = \frac{1}{\sin x} \).

Choose a Substitution

To evaluate this integral, recognize that the antiderivative of \( \csc x \) is not straightforward. Recall that the derivative of \( \ln|\csc x - \cot x| \) is \( \csc x \). So, the integral can be rewritten using this antiderivative.

Compute the Antiderivative

The antiderivative of \( \csc x \) is \( -\ln|\csc x + \cot x| \). Therefore, the evaluation for the indefinite integral is: \[ -\ln|\csc x + \cot x| + C \] where \( C \) is the constant of integration.

Evaluate the Definite Integral

Now evaluate the definite integral by substituting the limits \( \pi/6 \) and \( \pi/2 \):\[ -\ln|\csc(\pi/2) + \cot(\pi/2)| + \ln|\csc(\pi/6) + \cot(\pi/6)| \]

Calculate Specific Values

For \( x = \pi/2 \), \( \csc(\pi/2) = 1 \) and \( \cot(\pi/2) = 0 \). Therefore, \( \ln|\csc(\pi/2) + \cot(\pi/2)| = \ln|1 + 0| = 0 \).For \( x = \pi/6 \), \( \csc(\pi/6) = 2 \) and \( \cot(\pi/6) = \sqrt{3} \). Therefore, \( \ln|\csc(\pi/6) + \cot(\pi/6)| = \ln|2 + \sqrt{3}| \).

Final Calculation

Subtract the lower limit from the upper limit's result:\[ 0 - \ln|2 + \sqrt{3}| = -\ln|2 + \sqrt{3}| \].

Key concepts

Cosecant Function

The cosecant function, denoted as \( \csc x \), is one of the six main trigonometric functions. It is defined as the reciprocal of the sine function, so \( \csc x = \frac{1}{\sin x} \). This means wherever the sine of an angle is zero, the cosecant is undefined. This often occurs at integer multiples of \( \pi \).

Cosecant is particularly useful in trigonometric integration because it helps us relate problems involving right triangles to those involving circles. For example, in the unit circle where the radius is 1, \( \csc x \) gives the length of the hypotenuse over the opposite side, making it crucial for solving various integration problems.

Understanding \( \csc x \) is essential when dealing with integrals like \( \int \csc x \; dx \), as it often appears in contexts where sine cannot be easily used alone.

Antiderivative

An antiderivative is a function whose derivative results in the original function. For functions without a straightforward integration path, recognizing an antiderivative can save significant time.

In the exercise, we needed the antiderivative of \( \csc x \). Although this is not immediately intuitive, it is known from calculus textbooks that the antiderivative of \( \csc x \) is \( -\ln|\csc x + \cot x| + C \). Here, \( C \) stands for the constant of integration, appearing in indefinite integrals but omitting for definite integrals.

Knowing how to rewrite \( \csc x \) using its antiderivative form simplifies many trigonometric integrals. This approach is foundational in cases where partial fraction decomposition or further substitution might otherwise complicate the solution.

Trigonometric Integration

Trigonometric integration involves integrating functions that contain trigonometric expressions. It requires specific techniques and substitutions because many trigonometric functions don't have simple integrals.

In the provided solution, recognizing that the antiderivative of \( \csc x \) is required, leads us directly to using known results to evaluate this integral. Techniques such as substitution are invaluable in trigonometric integration, allowing us to transform complex functions into simpler ones.

Additionally, when integrating trigonometric functions, it helps to remember identities, like those expressing one function in terms of others (e.g., \( \sin x \) and \( \cos x \)). These often simplify expressions and make computing the integral more straightforward.

Limits of Integration

Limits of integration apply when dealing with definite integrals, providing boundaries for the area we are calculating under the curve of a function. In the exercise, the limits are \( \pi/6 \) and \( \pi/2 \), defining the interval over which \( \csc x \) is evaluated.

Evaluating a definite integral involves substituting these limits into the antiderivative and calculating the difference: \( F(b) - F(a) \). Here, \( F(x) \) represents the antiderivative, and \( a \) and \( b \) are the lower and upper limits, respectively.

In our case, calculating the integral within these limits involved evaluating the expression \( -\ln|\csc x + \cot x| \) at these specific points, resulting in \(-\ln|2 + \sqrt{3}|\). Understanding this process is central to mastering how to properly evaluate definite integrals.