Math Problem Statement

Integral((e^2x)*cos(2x)) from 0 to pi/2

Solution

To solve the integral 0π2e2xcos(2x)dx\int_0^{\frac{\pi}{2}} e^{2x} \cos(2x) \, dx, we will use integration by parts.

Recall that integration by parts is given by: udv=uvvdu\int u \, dv = uv - \int v \, du We need to choose uu and dvdv. Let's set: u=e2xanddv=cos(2x)dxu = e^{2x} \quad \text{and} \quad dv = \cos(2x) \, dx Then we need to find dudu and vv: du=2e2xdxandv=12sin(2x)du = 2e^{2x} \, dx \quad \text{and} \quad v = \frac{1}{2} \sin(2x)

Now we apply integration by parts: e2xcos(2x)dx=e2x12sin(2x)12sin(2x)2e2xdx\int e^{2x} \cos(2x) \, dx = e^{2x} \cdot \frac{1}{2} \sin(2x) - \int \frac{1}{2} \sin(2x) \cdot 2e^{2x} \, dx =12e2xsin(2x)e2xsin(2x)dx= \frac{1}{2} e^{2x} \sin(2x) - \int e^{2x} \sin(2x) \, dx

Now we need to integrate e2xsin(2x)dx\int e^{2x} \sin(2x) \, dx. We use integration by parts again for this integral. Set: u=e2xanddv=sin(2x)dxu = e^{2x} \quad \text{and} \quad dv = \sin(2x) \, dx Then: du=2e2xdxandv=12cos(2x)du = 2e^{2x} \, dx \quad \text{and} \quad v = -\frac{1}{2} \cos(2x)

Now applying integration by parts again: e2xsin(2x)dx=12e2xcos(2x)12cos(2x)2e2xdx\int e^{2x} \sin(2x) \, dx = -\frac{1}{2} e^{2x} \cos(2x) - \int -\frac{1}{2} \cos(2x) \cdot 2e^{2x} \, dx =12e2xcos(2x)+e2xcos(2x)dx= -\frac{1}{2} e^{2x} \cos(2x) + \int e^{2x} \cos(2x) \, dx

Notice that we end up with the original integral e2xcos(2x)dx\int e^{2x} \cos(2x) \, dx on the right-hand side. Let's call this integral II: I=e2xcos(2x)dxI = \int e^{2x} \cos(2x) \, dx So we have: I=12e2xsin(2x)(12e2xcos(2x))II = \frac{1}{2} e^{2x} \sin(2x) - \left( -\frac{1}{2} e^{2x} \cos(2x) \right) - I I=12e2xsin(2x)+12e2xcos(2x)II = \frac{1}{2} e^{2x} \sin(2x) + \frac{1}{2} e^{2x} \cos(2x) - I 2I=12e2xsin(2x)+12e2xcos(2x)2I = \frac{1}{2} e^{2x} \sin(2x) + \frac{1}{2} e^{2x} \cos(2x) I=14e2x(sin(2x)+cos(2x))I = \frac{1}{4} e^{2x} (\sin(2x) + \cos(2x))

Now we need to evaluate this expression from 00 to π2\frac{\pi}{2}: 14e2x(sin(2x)+cos(2x))0π2\left. \frac{1}{4} e^{2x} (\sin(2x) + \cos(2x)) \right|_0^{\frac{\pi}{2}} Evaluating at the bounds: 14e2x(sin(2x)+cos(2x))0π2=14e2π2(sin(π)+cos(π))14e20(sin(0)+cos(0))\left. \frac{1}{4} e^{2x} (\sin(2x) + \cos(2x)) \right|_0^{\frac{\pi}{2}} = \frac{1}{4} e^{2 \cdot \frac{\pi}{2}} (\sin(\pi) + \cos(\pi)) - \frac{1}{4} e^{2 \cdot 0} (\sin(0) + \cos(0)) =14eπ(0+(1))14(0+1)= \frac{1}{4} e^{\pi} (0 + (-1)) - \frac{1}{4} (0 + 1) =14eπ14= -\frac{1}{4} e^{\pi} - \frac{1}{4} =14(eπ+1)= -\frac{1}{4} (e^{\pi} + 1)

So the value of the integral is: 14(eπ+1)\boxed{-\frac{1}{4} (e^{\pi} + 1)}

Would you like more details or have any questions?

Here are some related questions:

  1. How do you integrate by parts?
  2. What are some common mistakes when applying integration by parts?
  3. How can we verify the result of an integral?
  4. What are the properties of exponential functions?
  5. How do trigonometric functions behave over different intervals?
  6. What are some applications of definite integrals?
  7. How can substitution simplify integrals?
  8. What are some techniques for integrating products of functions?

Tip: When using integration by parts, choosing uu and dvdv wisely can simplify the process significantly.

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Math Problem Analysis

Mathematical Concepts

Integration
Exponential Functions
Trigonometric Functions
Integration by Parts

Formulas

Integration by Parts Formula

Theorems

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Suitable Grade Level

Undergraduate Level