Math Problem Statement

Integrate sin4xcos^2(4x)

Solution

To integrate sin(4x)cos2(4x)\sin(4x) \cos^2(4x), we can simplify the expression by using trigonometric identities. Here's the step-by-step solution:

  1. Rewrite cos2(4x)\cos^2(4x) using the identity: cos2(4x)=1+cos(8x)2\cos^2(4x) = \frac{1 + \cos(8x)}{2} So the integral becomes: sin(4x)cos2(4x)dx=sin(4x)1+cos(8x)2dx\int \sin(4x) \cos^2(4x) \, dx = \int \sin(4x) \cdot \frac{1 + \cos(8x)}{2} \, dx

  2. Distribute sin(4x)\sin(4x) across the terms: =12sin(4x)dx+12sin(4x)cos(8x)dx= \frac{1}{2} \int \sin(4x) \, dx + \frac{1}{2} \int \sin(4x) \cos(8x) \, dx

  3. Integrate each term separately.

First Integral

For 12sin(4x)dx\frac{1}{2} \int \sin(4x) \, dx:

  • The antiderivative of sin(4x)\sin(4x) is 14cos(4x)-\frac{1}{4} \cos(4x), so: 12sin(4x)dx=12(14cos(4x))=18cos(4x)\frac{1}{2} \int \sin(4x) \, dx = \frac{1}{2} \cdot \left( -\frac{1}{4} \cos(4x) \right) = -\frac{1}{8} \cos(4x)

Second Integral

For 12sin(4x)cos(8x)dx\frac{1}{2} \int \sin(4x) \cos(8x) \, dx, we use the product-to-sum identity: sin(4x)cos(8x)=12(sin(12x)+sin(4x))\sin(4x) \cos(8x) = \frac{1}{2} \left( \sin(12x) + \sin(-4x) \right) Substituting, we get: 12sin(4x)cos(8x)dx=14(sin(12x)sin(4x))dx\frac{1}{2} \int \sin(4x) \cos(8x) \, dx = \frac{1}{4} \int \left( \sin(12x) - \sin(4x) \right) dx

Now we integrate each part separately:

  • The integral of sin(12x)\sin(12x) is 112cos(12x)-\frac{1}{12} \cos(12x).
  • The integral of sin(4x)-\sin(4x) is 14cos(4x)\frac{1}{4} \cos(4x).

Putting these together: 14(sin(12x)sin(4x))dx=14(112cos(12x)+14cos(4x))\frac{1}{4} \int \left( \sin(12x) - \sin(4x) \right) dx = \frac{1}{4} \left( -\frac{1}{12} \cos(12x) + \frac{1}{4} \cos(4x) \right) Simplifying, we get: =148cos(12x)+116cos(4x)= -\frac{1}{48} \cos(12x) + \frac{1}{16} \cos(4x)

Final Answer

Combining the results, we get: sin(4x)cos2(4x)dx=18cos(4x)148cos(12x)+116cos(4x)+C\int \sin(4x) \cos^2(4x) \, dx = -\frac{1}{8} \cos(4x) - \frac{1}{48} \cos(12x) + \frac{1}{16} \cos(4x) + C Simplify the terms involving cos(4x)\cos(4x): =148cos(12x)116cos(4x)+C= -\frac{1}{48} \cos(12x) - \frac{1}{16} \cos(4x) + C

Let me know if you’d like further clarification or details on any of these steps!


Additional Questions:

  1. How would you approach integrating functions involving higher powers of trigonometric functions?
  2. What are the product-to-sum identities, and how can they be applied to integration problems?
  3. How does simplifying trigonometric integrals benefit from using identities?
  4. What would the integration look like if we had sin2(4x)cos(4x)\sin^2(4x) \cos(4x) instead?
  5. What techniques are useful for integrating products of sine and cosine functions with different arguments?

Tip:

Applying trigonometric identities can often simplify complex integrals, making them easier to solve directly.

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Math Problem Analysis

Mathematical Concepts

Integration
Trigonometry
Trigonometric Identities
Product-to-Sum Identities

Formulas

cos^2(θ) = (1 + cos(2θ)) / 2
sin(A)cos(B) = (sin(A + B) + sin(A - B)) / 2
∫sin(kx) dx = -(1/k)cos(kx)

Theorems

Product-to-Sum Identity
Antiderivative Theorem

Suitable Grade Level

Grades 11-12