Math Problem Statement

integration by parts e^-x*cos(7x)dx

Solution

To solve the integral excos(7x)dx\int e^{-x} \cos(7x) \, dx using integration by parts, we will need to apply the method twice. Here are the steps:

Step 1: Set up integration by parts

Recall the integration by parts formula:

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

We choose:

  • u=cos(7x)u = \cos(7x), so du=7sin(7x)dxdu = -7 \sin(7x) \, dx
  • dv=exdxdv = e^{-x} \, dx, so v=exv = -e^{-x}

Step 2: Apply integration by parts

Now apply the formula:

excos(7x)dx=excos(7x)ex(7sin(7x))dx\int e^{-x} \cos(7x) \, dx = -e^{-x} \cos(7x) - \int -e^{-x} (-7 \sin(7x)) \, dx

Simplifying:

excos(7x)dx=excos(7x)+7exsin(7x)dx\int e^{-x} \cos(7x) \, dx = -e^{-x} \cos(7x) + 7 \int e^{-x} \sin(7x) \, dx

Step 3: Use integration by parts again

Now we need to solve the integral exsin(7x)dx\int e^{-x} \sin(7x) \, dx. We apply integration by parts again, this time choosing:

  • u=sin(7x)u = \sin(7x), so du=7cos(7x)dxdu = 7 \cos(7x) \, dx
  • dv=exdxdv = e^{-x} \, dx, so v=exv = -e^{-x}

Applying the formula again:

exsin(7x)dx=exsin(7x)ex(7cos(7x))dx\int e^{-x} \sin(7x) \, dx = -e^{-x} \sin(7x) - \int -e^{-x} (7 \cos(7x)) \, dx

Simplifying:

exsin(7x)dx=exsin(7x)7excos(7x)dx\int e^{-x} \sin(7x) \, dx = -e^{-x} \sin(7x) - 7 \int e^{-x} \cos(7x) \, dx

Step 4: Solve the system of equations

Now we have:

I=excos(7x)dxI = \int e^{-x} \cos(7x) \, dx and exsin(7x)dx=exsin(7x)7I\int e^{-x} \sin(7x) \, dx = -e^{-x} \sin(7x) - 7I

Substitute this back into the equation for the original integral:

I=excos(7x)+7(exsin(7x)7I)I = -e^{-x} \cos(7x) + 7(-e^{-x} \sin(7x) - 7I)

Simplify:

I=excos(7x)7exsin(7x)49II = -e^{-x} \cos(7x) - 7e^{-x} \sin(7x) - 49I

Step 5: Solve for II

Bring like terms together:

I+49I=ex(cos(7x)+7sin(7x))I + 49I = -e^{-x} (\cos(7x) + 7 \sin(7x))

50I=ex(cos(7x)+7sin(7x))50I = -e^{-x} (\cos(7x) + 7 \sin(7x))

Finally, solve for II:

I=ex(cos(7x)+7sin(7x))50I = \frac{-e^{-x} (\cos(7x) + 7 \sin(7x))}{50}

Thus, the solution to the integral is:

excos(7x)dx=ex(cos(7x)+7sin(7x))50+C\int e^{-x} \cos(7x) \, dx = \frac{-e^{-x} (\cos(7x) + 7 \sin(7x))}{50} + C

where CC is the constant of integration.


Would you like more details or have any questions?

Here are 5 related questions for further practice:

  1. How would you integrate e2xsin(5x)dx\int e^{2x} \sin(5x) \, dx?
  2. Can you apply integration by parts to xexdx\int x e^{-x} \, dx?
  3. How would you solve e3xcos(4x)dx\int e^{-3x} \cos(4x) \, dx?
  4. What is the integral of excos(2x)dx\int e^x \cos(2x) \, dx?
  5. How does the integration by parts formula relate to the product rule for differentiation?

Tip: Always check if repeating integration by parts will simplify the problem, especially with functions that involve products of exponential and trigonometric terms.

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Math Problem Analysis

Mathematical Concepts

Integration by Parts
Exponential Functions
Trigonometric Functions

Formulas

Integration by Parts: ∫ u dv = uv - ∫ v du
Trigonometric and Exponential Integration: Combining exponential and trigonometric integrals

Theorems

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Suitable Grade Level

Undergraduate Level (Calculus II)