Evaluate the following integral using integration by parts.
Integral from nothing to nothing x sine one seventh x dx
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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.

To evaluate the integral \(\int x \sin\left(\frac{1}{7}x\right) \, dx\) using integration by parts, we start by applying the formula for integration by parts:

\[
\int u \, dv = u v - \int v \, du
\]

### Step 1: Assign \( u \) and \( dv \)
We choose:
- \( u = x \), so \( du = dx \)
- \( dv = \sin\left(\frac{1}{7}x\right) \, dx \), so we need to find \( v \).

### Step 2: Compute \( v \)
We integrate \( dv = \sin\left(\frac{1}{7}x\right) \, dx \):

\[
v = \int \sin\left(\frac{1}{7}x\right) \, dx = -7 \cos\left(\frac{1}{7}x\right)
\]
(The factor of 7 comes from the chain rule when integrating.)

### Step 3: Apply the integration by parts formula
Now, we apply the formula:

\[
\int x \sin\left(\frac{1}{7}x\right) \, dx = u v - \int v \, du
\]

Substituting \( u = x \), \( du = dx \), \( v = -7 \cos\left(\frac{1}{7}x\right) \), we get:

\[
\int x \sin\left(\frac{1}{7}x\right) \, dx = -7x \cos\left(\frac{1}{7}x\right) - \int -7 \cos\left(\frac{1}{7}x\right) \, dx
\]

This simplifies to:

\[
\int x \sin\left(\frac{1}{7}x\right) \, dx = -7x \cos\left(\frac{1}{7}x\right) + 7 \int \cos\left(\frac{1}{7}x\right) \, dx
\]

### Step 4: Evaluate the remaining integral
We now need to compute \( \int \cos\left(\frac{1}{7}x\right) \, dx \):

\[
\int \cos\left(\frac{1}{7}x\right) \, dx = 7 \sin\left(\frac{1}{7}x\right)
\]

### Step 5: Final answer
Substituting this back into our equation:

\[
\int x \sin\left(\frac{1}{7}x\right) \, dx = -7x \cos\left(\frac{1}{7}x\right) + 49 \sin\left(\frac{1}{7}x\right) + C
\]

where \( C \) 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
\[
\boxed{-7x \cos\left(\frac{1}{7}x\right) + 49 \sin\left(\frac{1}{7}x\right) + C}
\]

Let me know if you need any further details or explanations!

#### 5 related questions:
1. How do you choose \( u \) and \( dv \) in integration by parts?
2. What is the chain rule in integration, and how does it apply here?
3. Can integration by parts be applied to trigonometric functions other than sine and cosine?
4. What happens when you apply integration by parts repeatedly?
5. How would this integral change if the limits of integration were specified?

**Tip:** When using integration by parts, always choose \( u \) as the part of the integrand that simplifies upon differentiation, and \( dv \) as the part that can be easily integrated.

Evaluate the following integral using integration by parts: Integral from nothing to nothing x sine one seventh x dx.

Learn how to solve the integral ∫x sin(1/7x) dx using the integration by parts method. This problem involves calculus concepts, including the integration by parts formula and trigonometric integrals. Suitable for advanced high school students or early college-level courses.

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