Question

For the following exercises, evaluate the definite integrals. Express answers in exact form whenever possible. \(\int_{0}^{2 \pi} \cos x \sin 2 x d x\)

Short answer

The integral evaluates to 0.

Step-by-step solution

Use Trigonometric Identity

Recall the trigonometric identity for the product of cosine and sine: \[ \cos x \sin 2x = \frac{1}{2} (\sin(3x) - \sin(x)) \].We'll substitute this expression into our integral.

Substitute and Simplify the Integral

Substitute the identity from Step 1 into the integral:\[ \int_{0}^{2\pi} \cos x \sin 2x \, dx = \int_{0}^{2\pi} \frac{1}{2} (\sin(3x) - \sin(x)) \, dx \]. This can be split into two separate integrals:\[ \frac{1}{2} \left( \int_{0}^{2\pi} \sin(3x) \, dx - \int_{0}^{2\pi} \sin(x) \, dx \right). \]

Evaluate the Integrals

The integral of \( \sin(x) \) over one full period \([0, 2\pi]\) is 0:\[ \int_{0}^{2\pi} \sin(x) \, dx = 0 \].Similarly, the integral of \( \sin(3x) \) over the same period is also 0:\[ \int_{0}^{2\pi} \sin(3x) \, dx = 0 \].When sin is integrated over its full cycle, the positive and negative areas cancel out.

Final Simplification

Substitute the evaluated integrals back into the expression:\[ \frac{1}{2} \left( 0 - 0 \right) = 0 \].Thus, the value of the definite integral is 0.

Key concepts

Trigonometric Identities

Trigonometric identities are essential tools in calculus, especially when dealing with integrals involving trigonometric functions. These identities help to simplify the expressions and make integration more feasible. In this exercise, we utilize the identity for the product of cosine and sine, specifically \( \cos x \sin 2x = \frac{1}{2} (\sin(3x) - \sin(x)) \). This identity allows us to rewrite the integral in a form that is easier to evaluate.

The purpose of using such identities is to transform complex products into simpler sums or differences of trigonometric functions. Once simplified, these functions can frequently be integrated directly. Learning and understanding these identities will streamline many integral problems, reduce the computational complexity, and often reveal symmetrical properties that are useful for solving definite integrals.

Integral Evaluation

Integral evaluation involves finding the value of an integral over a given interval. In the case of definite integrals, this evaluation results in a single numerical answer representing the signed area under the curve described by the function and above the x-axis, within the specified limits.

For this particular exercise, we simplify the integral \( \int_{0}^{2\pi} \cos x \sin 2x \, dx \) using the trigonometric identity transformed it into \( \frac{1}{2} \left( \int_{0}^{2\pi} \sin(3x) \, dx - \int_{0}^{2\pi} \sin(x) \, dx \right) \).

Each term is then evaluated over the interval from 0 to \( 2\pi \). Since both \( \int_{0}^{2\pi} \sin(x) \, dx \) and \( \int_{0}^{2\pi} \sin(3x) \, dx \) result in zero (because these functions complete their full cycles), the result of the integral is zero. This step-by-step process highlights how simplification and substitution are key in evaluating integrals effectively.

Periodic Functions

Periodic functions repeatedly oscillate through a regular pattern over set intervals, characterized by their period. In trigonometry, sine and cosine functions are perfect examples of periodic functions. Their periodicity has significant implications when evaluating integrals over their full period.

For the functions involved in this problem, \( \sin(x) \) and \( \sin(3x) \), their integrals over the interval from 0 to \( 2\pi \) (a full period for both) equal zero.

• Since sine functions are odd, they symmetrically mirror across the x-axis.
• This symmetry means that the area above the axis cancels out the area below.

The nature of periodic functions often leads to simplifications in integral calculus, particularly for definite integrals over a complete cycle. Understanding this property allows for more effective problem solving and highlights the elegant mathematical symmetry often encountered in calculus.