Math Problem Statement
Solution
The image contains a series of integration problems (4.1 to 4.5). Let me solve them step by step:
4.1.1
To integrate :
- Use the identity: .
- Let . Then .
Substitute:
Final result:
4.1.2
Use the product-to-sum formula:
So: [ \sin(-6x)\sin(3x) = \frac{1}{2}[\cos(-6x - 3x) - \cos(-6x + 3x)] = \frac{1}{2}[\cos(-9x) - \cos(-3x)]. ]
Now integrate term-by-term:
Substitute:
Final result:
4.1.3
Use the identity:
Let . Then:
Integrate term-by-term:
Final result:
4.1.4
This requires polynomial long division:
- Perform division: Divide by .
- Integrate the result.
4.1.5
Here, the negative sign in doesn't change the result ():
Use integration by parts: Let:
Then:
Integration by parts formula:
Substitute:
Solve recursively.
Let me know if you want solutions for specific cases, or for me to detail partial fraction decomposition for . 😊
Follow-up Questions
- Do you want further breakdown of integration techniques like substitution or parts?
- Should I explain polynomial long division for ?
- Would you like to see the detailed steps for (recursive integration)?
- Do you need help with simplifying trigonometric identities?
- Are there any specific steps where you'd like clarification?
Tip:
Always check if trigonometric integrals can be simplified using identities before proceeding to solve. It often makes the process faster!
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Math Problem Analysis
Mathematical Concepts
Trigonometric identities
Integration by substitution
Integration by parts
Partial fractions decomposition
Polynomial long division
Formulas
∫sec²(u)tan(u) du = sec²(u)
sin(A)sin(B) = 1/2[cos(A-B) - cos(A+B)]
sin²(u) = 1/2[1 - cos(2u)]
∫cos(kx) dx = sin(kx)/k
∫u dv = uv - ∫v du
Theorems
Trigonometric product-to-sum identities
Integration by parts theorem
Fundamental theorem of calculus
Suitable Grade Level
College-level Calculus
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