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 xsin(17x)dx\int x \sin\left(\frac{1}{7}x\right) \, dx using integration by parts, we start by applying the formula for integration by parts:

udv=uvvdu\int u \, dv = u v - \int v \, du

Step 1: Assign uu and dvdv

We choose:

  • u=xu = x, so du=dxdu = dx
  • dv=sin(17x)dxdv = \sin\left(\frac{1}{7}x\right) \, dx, so we need to find vv.

Step 2: Compute vv

We integrate dv=sin(17x)dxdv = \sin\left(\frac{1}{7}x\right) \, dx:

v=sin(17x)dx=7cos(17x)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:

xsin(17x)dx=uvvdu\int x \sin\left(\frac{1}{7}x\right) \, dx = u v - \int v \, du

Substituting u=xu = x, du=dxdu = dx, v=7cos(17x)v = -7 \cos\left(\frac{1}{7}x\right), we get:

xsin(17x)dx=7xcos(17x)7cos(17x)dx\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:

xsin(17x)dx=7xcos(17x)+7cos(17x)dx\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 cos(17x)dx\int \cos\left(\frac{1}{7}x\right) \, dx:

cos(17x)dx=7sin(17x)\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:

xsin(17x)dx=7xcos(17x)+49sin(17x)+C\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 CC 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

7xcos(17x)+49sin(17x)+C\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 uu and dvdv 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 uu as the part of the integrand that simplifies upon differentiation, and dvdv 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