Question

\(\int \sin x \cdot \cos x \cdot \cos 2 x \cdot \cos 4 x \cdot \cos 8 x \cdot \cos 16 x d x=\ldots+c\) (a) \([(\sin 16 \mathrm{x}) /(1024)]\) (b) \(-[(\cos 32 \mathrm{x}) /(1024)]\) (c) \([(\cos 32 \mathrm{x}) /(1096)]\) (d) \(-[(\cos 32 \mathrm{x}) /(1096)]\)

Short answer

The correct answer is (b) \(\frac{\cos(32x)}{1024}+C\).

Step-by-step solution

Apply product-to-sum identities

We notice that the integrand has multiple sine and cosine terms. To simplify the expression, we can use the product-to-sum identities: \(\sin(u)\cos(v) = \frac{1}{2}[\sin(u - v) - \sin(u + v)]\) Let's apply this identity to \(\sin(x)\cos(x)\), resulting in: \(\frac{1}{2}[\sin(x-x)- \sin(x+x)] = \frac{1}{2}[-\sin(2x)]\) Now the function we need to integrate becomes: \(-\frac{1}{2}\sin(2x)\cos(2x)\cos(4x)\cos(8x)\cos(16x)\)

Apply product-to-sum identities again

Since we still have products of trigonometric functions, we can apply the product-to-sum identities again to \(\sin(2x)\cos(2x)\): \(-\frac{1}{4}[\sin(4x) - \sin(0x)] = -\frac{1}{4}\sin(4x)\) Now the function we need to integrate becomes: \(-\frac{1}{4}\sin(4x)\cos(4x)\cos(8x)\cos(16x)\)

Apply product-to-sum identities for the last time

We apply the product-to-sum identities again to \(\sin(4x)\cos(4x)\): \(-\frac{1}{8}[\sin(8x) - \sin(0x)] = -\frac{1}{8}\sin(8x)\) Now the function we need to integrate becomes: \(-\frac{1}{8}\sin(8x)\cos(8x)\cos(16x)\)

Integrate the simplified expression

We can now integrate the simplified expression: \(-\frac{1}{8}\int\sin(8x)\cos(8x)\cos(16x) dx\) Once again, we apply the product-to-sum identities to \(\sin(8x)\cos(8x)\): \(-\frac{1}{16}\int[\sin(16x) - \sin(0x)]\cos(16x) dx = -\frac{1}{16}\int\sin(16x)\cos(16x)dx\) Now, we can apply the identity one last time, to \(\sin(16x)\cos(16x)\): \(-\frac{1}{32}\int[\sin(32x) - \sin(0x)] dx = -\frac{1}{32}\int\sin(32x)dx\) Finally, integrating the remaining expression: \(-\frac{1}{32}\int\sin(32x)dx = -\frac{1}{32}(-\frac{1}{32}\cos(32x)) + C = \frac{1}{1024}\cos(32x) +C\) This matches option (b), so the answer is: \(\boxed{\text{(b)}\ \frac{\cos(32x)}{1024}+C}\)

Key concepts

Product-to-Sum Identities

Mastering product-to-sum identities is a powerful tool in the world of trigonometry. These identities allow us to transform products of sine and cosine functions into sums or differences, facilitating simpler integration or differentiation tasks. In the given exercise, understanding how to use product-to-sum identities was vital in breaking down the complex integrand.

Product-to-sum identities take the form \[ \(\sin(u)\cos(v) = \frac{1}{2}[\sin(u - v) - \sin(u + v)]\).\] Using these identities, we can simplify products like \(\sin(x)\cos(x)\) into a single sine function that is much easier to integrate. This approach was repeated at each step of the solution to continuously simplify the problem until it was possible to directly integrate the final trigonometric function.

Integration Techniques

The world of calculus is filled with various integration techniques that make solving complex integrals feasible. These range from simple methods, like the power rule, to more advanced strategies, like u-substitution, integration by parts, trigonometric substitution, and partial fraction decomposition. Each technique has its place, and knowing when to use them is key to solving integrals effectively.

In our textbook example, it was essential to recognize that a product-to-sum identity could simplify the integrand before applying direct integration. Understanding how and when to use a specific method can transform an otherwise daunting integral into a manageable one, as demonstrated.

Trigonometric Integration

Trigonometric integration involves finding the antiderivative of functions involving sine, cosine, tangent, secant, and other trigonometric functions. There are numerous strategies for handling these types of integrals, such as using trigonometric identities, converting to exponential forms, or applying trig-specific substitution rules.

What makes trigonometric integration interesting—and sometimes challenging—is that it often requires a combination of techniques to achieve the solution. For instance, the exercise provided not only utilized product-to-sum identities multiple times but also showcased how the integration process often involves a step-by-step simplification of the function being integrated. Each simplification brought the function closer to a form that could finally be integrated straightforwardly to find the answer.