Math Problem Statement

\int :e^t\sin \left(2t\right)du

Solution

The integral you've provided seems to have a slight issue with the notation. It should likely be:

etsin(2t)dt\int e^t \sin(2t) \, dt

Here, tt is the variable of integration. Assuming this is correct, let's solve this integral using the method of integration by parts.

The integral etsin(2t)dt\int e^t \sin(2t) \, dt can be solved by using integration by parts or using a tabular method.

Solution using integration by parts:

We can apply the formula for integration by parts twice to solve this integral.

Let: I=etsin(2t)dtI = \int e^t \sin(2t) \, dt

  1. Use integration by parts. Set u=sin(2t)u = \sin(2t) and dv=etdtdv = e^t \, dt.
    • du=2cos(2t)dtdu = 2\cos(2t) \, dt
    • v=etv = e^t

Applying the integration by parts formula:

I=etsin(2t)et2cos(2t)dtI = e^t \sin(2t) - \int e^t \cdot 2 \cos(2t) \, dt

  1. Now, we need to solve the new integral etcos(2t)dt\int e^t \cos(2t) \, dt, which requires integration by parts again. Let:

    • u=cos(2t)u = \cos(2t)
    • dv=etdtdv = e^t \, dt

    Then,

    • du=2sin(2t)dtdu = -2\sin(2t) \, dt
    • v=etv = e^t

Now applying integration by parts again:

etcos(2t)dt=etcos(2t)et(2sin(2t))dt\int e^t \cos(2t) \, dt = e^t \cos(2t) - \int e^t \cdot (-2\sin(2t)) \, dt etcos(2t)dt=etcos(2t)+2etsin(2t)dt\int e^t \cos(2t) \, dt = e^t \cos(2t) + 2 \int e^t \sin(2t) \, dt

  1. Substituting back into the original equation: I=etsin(2t)2(etcos(2t)+2etsin(2t)dt)I = e^t \sin(2t) - 2 \left( e^t \cos(2t) + 2 \int e^t \sin(2t) \, dt \right)

Now, simplify and solve for II: I=etsin(2t)2etcos(2t)4II = e^t \sin(2t) - 2e^t \cos(2t) - 4I

  1. Combine like terms: I+4I=etsin(2t)2etcos(2t)I + 4I = e^t \sin(2t) - 2e^t \cos(2t) 5I=et(sin(2t)2cos(2t))5I = e^t (\sin(2t) - 2\cos(2t)) I=et(sin(2t)2cos(2t))5+CI = \frac{e^t (\sin(2t) - 2\cos(2t))}{5} + C

Thus, the solution is:

etsin(2t)dt=et(sin(2t)2cos(2t))5+C\int e^t \sin(2t) \, dt = \frac{e^t (\sin(2t) - 2\cos(2t))}{5} + C

Do you need any further clarification or details on the steps?

Related questions:

  1. How does integration by parts work in general?
  2. What is the tabular method for solving integrals like this?
  3. Can integration by parts be used for any type of integral?
  4. How do we derive the formula for integration by parts?
  5. What are the differences between using integration by parts and substitution?

Tip:

When using integration by parts repeatedly, it's useful to recognize patterns or use a tabular method to avoid excessive computation.

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Math Problem Analysis

Mathematical Concepts

Integration
Trigonometric Functions
Exponential Functions
Integration by Parts

Formulas

Integration by parts: \int u dv = uv - \int v du
Trigonometric identities: \sin(2t), \cos(2t)

Theorems

Integration by Parts Theorem

Suitable Grade Level

Undergraduate Calculus (Grades 11-12 or College Level)