Math Problem Statement
Evaluate the following integral using integration by parts. Integral from nothing to nothing x sine one seventh x dx Question content area bottom Part 1 Use the integration by parts formula so that the new integral is simpler than the original one. Choose the correct answer below. A.negative 7 cosine one seventh x minus Integral from nothing to nothing left parenthesis negative 7 x cosine one seventh x right parenthesis dx negative 7 cosine one seventh x minus Integral from nothing to nothing left parenthesis negative 7 x cosine one seventh x right parenthesis dx B.one seventh x cosine 7 x plus Integral from nothing to nothing left parenthesis one seventh cosine 7 x right parenthesis dx one seventh x cosine 7 x plus Integral from nothing to nothing left parenthesis one seventh cosine 7 x right parenthesis dx C.7 x cosine one seventh x plus Integral from nothing to nothing left parenthesis 7 cosine one seventh x right parenthesis dx 7 x cosine one seventh x plus Integral from nothing to nothing left parenthesis 7 cosine one seventh x right parenthesis dx D.negative 7 x cosine one seventh x minus Integral from nothing to nothing left parenthesis negative 7 cosine one seventh x right parenthesis dx negative 7 x cosine one seventh x minus Integral from nothing to nothing left parenthesis negative 7 cosine one seventh x right parenthesis dx Part 2 Evaluate the integral.
Solution
To evaluate the integral using integration by parts, we start by applying the formula for integration by parts:
Step 1: Assign and
We choose:
- , so
- , so we need to find .
Step 2: Compute
We integrate :
(The factor of 7 comes from the chain rule when integrating.)
Step 3: Apply the integration by parts formula
Now, we apply the formula:
Substituting , , , we get:
This simplifies to:
Step 4: Evaluate the remaining integral
We now need to compute :
Step 5: Final answer
Substituting this back into our equation:
where is the constant of integration.
Part 1: Choosing the correct answer
Among the choices given:
- The correct answer corresponds to Option D, because it matches the structure of the integration by parts we performed.
Part 2: The final evaluated integral
Let me know if you need any further details or explanations!
5 related questions:
- How do you choose and in integration by parts?
- What is the chain rule in integration, and how does it apply here?
- Can integration by parts be applied to trigonometric functions other than sine and cosine?
- What happens when you apply integration by parts repeatedly?
- How would this integral change if the limits of integration were specified?
Tip: When using integration by parts, always choose as the part of the integrand that simplifies upon differentiation, and as the part that can be easily integrated.
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Math Problem Analysis
Mathematical Concepts
Calculus
Integration by Parts
Trigonometric Functions
Formulas
Integration by Parts formula: ∫u dv = uv - ∫v du
Trigonometric integrals: ∫sin(ax) dx = -1/a cos(ax), ∫cos(ax) dx = 1/a sin(ax)
Theorems
Integration by Parts Theorem
Chain Rule for Integration
Suitable Grade Level
Grades 11-12 or early college level