Math Problem Statement
\int :sin^4\left(x\right):cos^2:\left(x\right)
Solution
To evaluate the integral , we can simplify the integrand using trigonometric identities.
Step 1: Use power-reducing identities
We start by converting powers of sine and cosine using the following identities:
\quad \text{and} \quad \cos^2(x) = \frac{1 + \cos(2x)}{2}.$$ Thus, we rewrite $$\sin^4(x) \cos^2(x)$$ as: $$\sin^4(x) = (\sin^2(x))^2 = \left( \frac{1 - \cos(2x)}{2} \right)^2,$$ so the integrand becomes: $$\int \left( \frac{1 - \cos(2x)}{2} \right)^2 \cdot \frac{1 + \cos(2x)}{2} \, dx.$$ ### Step 2: Expand the product First, expand the square $$\left( \frac{1 - \cos(2x)}{2} \right)^2$$: $$\left( \frac{1 - \cos(2x)}{2} \right)^2 = \frac{1 - 2\cos(2x) + \cos^2(2x)}{4}.$$ Now multiply this by $$\frac{1 + \cos(2x)}{2}$$: $$\frac{1 - 2\cos(2x) + \cos^2(2x)}{4} \cdot \frac{1 + \cos(2x)}{2} = \frac{1}{8} \left( (1 - 2\cos(2x) + \cos^2(2x))(1 + \cos(2x)) \right).$$ ### Step 3: Expand further Now expand the expression inside the parentheses: $$(1 - 2\cos(2x) + \cos^2(2x))(1 + \cos(2x)) = 1 + \cos(2x) - 2\cos(2x) - 2\cos^2(2x) + \cos^2(2x) + \cos^3(2x).$$ Simplify this: $$1 - \cos(2x) - \cos^2(2x) + \cos^3(2x).$$ ### Step 4: Split the integral Now the integrand is: $$\frac{1}{8} \int \left( 1 - \cos(2x) - \cos^2(2x) + \cos^3(2x) \right) dx.$$ We can split this into four simpler integrals: $$\frac{1}{8} \left( \int 1 \, dx - \int \cos(2x) \, dx - \int \cos^2(2x) \, dx + \int \cos^3(2x) \, dx \right).$$ ### Step 5: Evaluate each integral 1. $$\int 1 \, dx = x$$. 2. $$\int \cos(2x) \, dx = \frac{\sin(2x)}{2}$$. 3. $$\int \cos^2(2x) \, dx$$ uses the power-reducing identity $$\cos^2(2x) = \frac{1 + \cos(4x)}{2}$$, so: $$\int \cos^2(2x) \, dx = \frac{1}{2} \int 1 + \cos(4x) \, dx = \frac{x}{2} + \frac{\sin(4x)}{8}.$$ 4. $$\int \cos^3(2x) \, dx$$ uses the identity $$\cos^3(2x) = \cos(2x)(\frac{1 + \cos(4x)}{2})$$, which can be expanded and integrated term by term. However, this step involves a bit more algebra, and the result is: $$\int \cos^3(2x) \, dx = \frac{3 \sin(2x)}{8} + \frac{\sin(6x)}{48}.$$ ### Final Answer: The full solution is: $$\frac{1}{8} \left( x - \frac{\sin(2x)}{2} - \left( \frac{x}{2} + \frac{\sin(4x)}{8} \right) + \left( \frac{3\sin(2x)}{8} + \frac{\sin(6x)}{48} \right) \right) + C.$$ This can be simplified further, but this is the general structure. Would you like more details or have any questions? --- Here are 5 related questions: 1. How do you apply trigonometric identities to simplify integrals? 2. What is the power-reducing identity for higher powers of sine and cosine? 3. How do you handle integrals involving products of trigonometric functions? 4. What are some strategies for solving integrals of powers of trigonometric functions? 5. How do you simplify the integration of cubic trigonometric functions? **Tip:** Using symmetry in trigonometric functions can simplify integrals, especially when the powers are even.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Trigonometric Integrals
Power-Reducing Identities
Trigonometric Expansions
Formulas
\sin^2(x) = \frac{1 - \cos(2x)}{2}
\cos^2(x) = \frac{1 + \cos(2x)}{2}
Power-reducing and product formulas for trigonometric functions
Theorems
Power-Reducing Identity
Trigonometric Expansion
Integral Properties for Trigonometric Functions
Suitable Grade Level
Undergraduate Mathematics or Advanced High School Calculus
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