Question

Evaluate the definite integral. $$ \int_{\pi / 6}^{\pi / 3} \csc x \cot x d x $$

Short answer

\( 2 - \frac{2\sqrt{3}}{3} \)

Step-by-step solution

Understand the Function to Integrate

The integral given is \( \int_{\pi / 6}^{\pi / 3} \csc x \cot x \, dx \). The function \( \csc x \cot x \) is a product of two trigonometric functions, cosecant and cotangent.

Know the Antiderivative

Recall that the derivative of \(-\csc x\) is \( \csc x \cot x \). Thus, the antiderivative of \( \csc x \cot x \) is \(-\csc x\).

Apply the Fundamental Theorem of Calculus

Using the antiderivative from Step 2, evaluate the integral from \( \pi / 6 \) to \( \pi / 3 \):\[ \int_{\pi / 6}^{\pi / 3} \csc x \cot x \, dx = \left[ - \csc x \right]_{\pi / 6}^{\pi / 3} \]

Evaluate the Boundaries

Substitute the limits of integration into the antiderivative:\[ -\csc(\pi/3) - (-\csc(\pi/6)) = -\csc(\pi/3) + \csc(\pi/6) \]

Calculate Using Known Values

Know that \( \csc(\pi/3) = \frac{2}{\sqrt{3}} \) and \( \csc(\pi/6) = 2 \). Substitute these values:\[ -\frac{2}{\sqrt{3}} + 2 \]

Simplify the Result

To simplify:\[ 2 - \frac{2}{\sqrt{3}} \] Rationalize the denominator:\[ 2 - \frac{2\sqrt{3}}{3} \] Thus, the final simplified answer is \( 2 - \frac{2\sqrt{3}}{3} \).

Key concepts

trigonometric functions

Trigonometric functions are fundamental in the study of mathematics, especially in calculus and trigonometry. In this exercise, we encounter two specific trigonometric functions: cosecant (\( \csc x \)) and cotangent (\( \cot x \)). Both functions are related to the more commonly known sine and cosine functions.

• Cosecant is the reciprocal of the sine function, defined as \( \csc x = \frac{1}{\sin x} \).
• Cotangent is the reciprocal of the tangent function, defined as \( \cot x = \frac{\cos x}{\sin x} \).

These functions are useful when dealing with right triangles and cyclic phenomena. In calculus, they often appear in integration and differentiation processes. Understanding how to manipulate these trigonometric functions is essential for successfully evaluating integrals like the one in the original problem.

antiderivative

When solving integrals, the antiderivative plays a critical role. Finding an antiderivative involves reversing the process of differentiation. For the integral in the exercise, the goal was to find an antiderivative of the trigonometric expression \( \csc x \cot x \).
The important thing to remember is that if you know the derivative of a function, you can identify its antiderivative. In this case, the derivative of \(-\csc x\) is \( \csc x \cot x \). Thus, the antiderivative of \( \csc x \cot x \) is \(-\csc x\).
Recognizing derivative patterns quickly aids in solving problems efficiently. This skill is particularly important in calculus when dealing with indefinite and definite integrals.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a cornerstone of modern calculus. It connects differentiation and integration, providing a way to evaluate definite integrals efficiently. For the integration problem in the exercise, this theorem was used to simplify the computation.

• The theorem states that if \( F \) is an antiderivative of \( f \) over an interval, then: \[ \int_{a}^{b} f(x) \, dx = F(b) - F(a) \]

By applying this concept, we calculated the definite integral of \( \csc x \cot x \) from \( \pi/6 \) to \( \pi/3 \). We substituted these upper and lower bounds into the antiderivative \(-\csc x\). Understanding and using this theorem effectively can greatly ease the process of solving integrals, particularly when dealing with complex functions.