Question

For the following exercises, evaluate the definite integrals. Express answers in exact form whenever possible. \(\int_{0}^{2 \pi} \sin x \sin (2 x) \sin (3 x) d x\)

Short answer

The integral evaluates to 0.

Step-by-step solution

Apply Product-to-Sum Formula

To evaluate the integral \( \int_{0}^{2 \pi} \sin x \sin (2 x) \sin (3 x) \, dx \), we start by simplifying the integrand using the product-to-sum identities. The identity for product-to-sum for two sines is \( \sin A \sin B = \frac{1}{2} [\cos(A-B) - \cos(A+B)] \). Let's first simplify \( \sin x \sin (2x) \):\[ \sin x \sin (2x) = \frac{1}{2}[\cos(x - 2x) - \cos(x + 2x)] = \frac{1}{2}[\cos(-x) - \cos(3x)] = \frac{1}{2}[\cos x - \cos 3x] \].Now express the remaining expression using this identity:\( \frac{1}{2} [\cos x - \cos 3x] \sin (3x) \). You will use product-to-sum again for each product term.

Further Simplification Using Product-to-Sum

Now apply the product-to-sum identity to \( \cos x \sin 3x \) and \( \cos 3x \sin 3x \):1. \( \cos x \sin 3x = \frac{1}{2} [\sin(3x + x) - \sin(3x - x)] = \frac{1}{2} [\sin 4x - \sin 2x] \).2. \( \cos 3x \sin 3x = \frac{1}{2} [\sin(3x+3x) - \sin(3x-3x)] = \frac{1}{2} [\sin 6x - 0] = \frac{1}{2} \sin 6x \).Substitute back into the expression:\[ \frac{1}{4} [\sin 4x - \sin 2x] - \frac{1}{4} [\sin 6x] \].Thus, the integrand becomes:\[ \frac{1}{4} \sin 4x - \frac{1}{4} \sin 2x - \frac{1}{4} \sin 6x \].

Integrate Each Term Separately

Now, integrate each term separately between the limits 0 and \( 2 \pi \):1. \( \int_{0}^{2\pi} \frac{1}{4} \sin 4x \, dx = \frac{1}{4} \left[ -\frac{1}{4} \cos 4x \right]_{0}^{2\pi} = \frac{-1}{16} \left[ \cos(8\pi) - \cos(0) \right] = 0 \) because \( \cos(8 \pi) = \cos(0) = 1 \).2. \( \int_{0}^{2\pi} \frac{1}{4} \sin 2x \, dx = \frac{1}{4} \left[ -\frac{1}{2} \cos 2x \right]_{0}^{2\pi} = \frac{-1}{8} \left[ \cos(4\pi) - \cos(0) \right] = 0 \) because \( \cos(4 \pi) = \cos(0) = 1 \).3. \( \int_{0}^{2\pi} \frac{1}{4} \sin 6x \, dx = \frac{1}{4} \left[ -\frac{1}{6} \cos 6x \right]_{0}^{2\pi} = \frac{-1}{24} \left[ \cos(12\pi) - \cos(0) \right] = 0 \) because \( \cos(12 \pi) = \cos(0) = 1 \).Thus, all integrals from 0 to \(2\pi\) of each term are zero.

Sum the Results

After integrating each term, add the results together:\( 0 + 0 + 0 = 0 \).So, the integral \( \int_{0}^{2\pi} \sin x \sin (2x) \sin(3x) \, dx = 0 \).

Key concepts

Product-to-Sum Identities

Product-to-sum identities are invaluable tools in trigonometry, especially when you encounter products of sine and cosine functions in integrals. The identities transform these products into sums or differences, which are often easier to integrate.

A basic example is the identity for the product of two sine functions:

• \(\sin A \sin B = \frac{1}{2} [\cos(A-B) - \cos(A+B)]\).

First, it's crucial to apply these identities correctly. In the given exercise, we started by simplifying \(\sin x \sin (2x)\) using the product-to-sum formula. It leads to:

• \(\frac{1}{2} [\cos x - \cos 3x]\).

This simplification helps to break down complex trigonometric functions into manageable pieces, turning multiplication into addition or subtraction, thereby making the integrals simpler.

Remember to apply the formula step-by-step, breaking down each product within the integral, as illustrated in the solution, to ensure no part of the original expression is overlooked. This initial simplification sets the stage for ultimately solving the integral.

Trigonometric Integrals

Trigonometric integrals ask us to find the integral of trigonometric functions. They often require special techniques due to their oscillating nature. In dealing with such integrals, simplication strategies or known identities, like product-to-sum, are frequently used.

For the given integral \( \int_{0}^{2 \pi} \sin x \sin (2x) \sin (3x) \, dx \), simplifying each component using identities allows for easier integration. After transforming the product into simpler terms, we integrate terms like \( \sin 4x\), \( \sin 2x\), and \( \sin 6x\).

These simpler terms, while still trigonometric, can be integrated straightforwardly using basic antiderivatives:

• The antiderivative of \( \sin(kx) \) is \( -\frac{1}{k} \cos(kx) \).

The definite integral of sine over a full period (like from 0 to \( 2\pi\)) results in zero because the areas above and below the x-axis cancel out over one full cycle. Thus, correctly applying these identities and recognizing these cycles plays a key role in solving the integral successfully.

Integration by Parts

Integration by parts is a technique derived from the product rule for differentiation and is particularly useful for integrating products of functions. It is formulated as:

• \(\int u \, dv = uv - \int v \, du\).

This formula helps to transform the integral into a simpler one by choosing parts of the expression to set as \( u \) and \( dv \). Although not used directly in the main steps of this exercise, understanding integration by parts provides great depth in tackling more complex integrals involving products of functions.

When working with multiple functions, try to choose \( u\) to be something that simplifies when differentiated, and \( dv\) as something easily integrable. This strategic approach simplifies the problem iteratively.

In contexts where trigonometric integrals become more intricate, the combination of product-to-sum identities and integration by parts can be highly effective. Recognizing when each method is applicable is key to mastering calculus, especially as encountered in this exercise with trigonometric products.