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Integrate the following problem:
[tex]\int {e^{-x} \cdot cos(2x)} \, dx[/tex]

P.S:
Have fun Lauren (and possibly Kelvy)!

Respuesta :

Supernova

Answer:

[tex]\displaystyle \frac{2 \cdot sin2x-cos2x}{5e^x} + C[/tex]

Step-by-step explanation:

The integration by parts formula is: [tex]\displaystyle \int udv = uv - \int vdu[/tex]

Let's find u, du, dv, and v for [tex]\displaystyle \int e^-^x \cdot cos2x \ dx[/tex] .

  • [tex]u=e^-^x[/tex]
  • [tex]du=-e^-^x dx[/tex]
  • [tex]dv=cos2x \ dx[/tex]
  • [tex]v= \frac{sin2x}{2}[/tex]

Plug these values into the IBP formula:

  • [tex]\displaystyle \int e^-^x \cdot cos2x \ dx = e^-^x \cdot \frac{sin2x}{2} - \int \frac{sin2x}{2} \cdot -e^-^x dx[/tex]
  • [tex]\displaystyle \int e^-^x \cdot cos2x \ dx = \frac{e^-^x sin2x}{2} - \int \frac{sin2x}{2} \cdot -e^-^x dx[/tex]

Now let's evaluate the integral [tex]\displaystyle \int \frac{sin2x}{2} \cdot -e^-^x dx[/tex].

Let's find u, du, dv, and v for this integral:

  • [tex]u=-e^-^x[/tex]
  • [tex]du=e^-^x dx[/tex]
  • [tex]dv=\frac{sin2x}{2} dx[/tex]
  • [tex]v=\frac{-cos2x}{4}[/tex]  

Plug these values into the IBP formula:

  • [tex]\displaystyle \int -e^-^x \cdot \frac{sin2x}{x}dx = -e^-^x \cdot \frac{-cos2x}{4} - \int \frac{-cos2x}{4}\cdot e^-^x dx[/tex]

Factor 1/4 out of the integral and we are left with the exact same integral from the question.

  • [tex]\displaystyle \int -e^-^x \cdot \frac{sin2x}{x}dx = -e^-^x \cdot \frac{-cos2x}{4} + \frac{1}{4} \int cos2x \cdot e^-^x dx[/tex]

Let's substitute this back into the first IBP equation.

  • [tex]\displaystyle \int e^-^x \cdot cos2x \ dx = \frac{e^-^x sin2x}{2} - \Big [ -e^-^x \cdot \frac{-cos2x}{4} + \frac{1}{4} \int cos2x \cdot e^-^x dx \Big ][/tex]  

Simplify inside the brackets.

  • [tex]\displaystyle \int e^-^x \cdot cos2x \ dx = \frac{e^-^x sin2x}{2} - \Big [ \frac{e^-^x \cdot cos2x}{4} + \frac{1}{4} \int cos2x \cdot e^-^x dx \Big ][/tex]

Distribute the negative sign into the parentheses.

  • [tex]\displaystyle \int e^-^x \cdot cos2x \ dx = \frac{e^-^x sin2x}{2} - \frac{e^-^x \cdot cos2x}{4} - \frac{1}{4} \int cos2x \cdot e^-^x dx[/tex]

Add the like term to the left side.

  • [tex]\displaystyle \int e^-^x \cdot cos2x \ dx + \frac{1}{4} \int cos2x \cdot e^-^x dx= \frac{e^-^x sin2x}{2} - \frac{e^-^x \cdot cos2x}{4}[/tex]  
  • [tex]\displaystyle \frac{5}{4} \int e^-^x \cdot cos2x \ dx = \frac{e^-^x sin2x}{2} - \frac{e^-^x \cdot cos2x}{4}[/tex]  

Make the fractions have common denominators.

  • [tex]\displaystyle \frac{5}{4} \int e^-^x \cdot cos2x \ dx = \frac{2e^-^x sin2x}{4} - \frac{e^-^x \cdot cos2x}{4}[/tex]

Simplify this equation.

  • [tex]\displaystyle \frac{5}{4} \int e^-^x \cdot cos2x \ dx = \frac{2e^-^x sin2x - e^-^x cos2x}{4}[/tex]

Multiply the right side by the reciprocal of 5/4.

  • [tex]\displaystyle \int e^-^x \cdot cos2x \ dx = \frac{2e^-^x sin2x - e^-^x cos2x}{4} \cdot \frac{4}{5}[/tex]

The 4's cancel out and we are left with:

  • [tex]\displaystyle \int e^-^x \cdot cos2x \ dx = \frac{2e^-^x sin2x - e^-^x cos2x}{5}[/tex]

Factor [tex]e^-^x[/tex] out of the numerator.

  • [tex]\displaystyle \int e^-^x \cdot cos2x \ dx = \frac{e^-^x(2 \cdot sin2x-cos2x)}{5}[/tex]

Simplify this by using exponential properties.

  • [tex]\displaystyle \int e^-^x \cdot cos2x \ dx = \frac{2 \cdot sin2x-cos2x}{5e^x}[/tex]

The final answer is [tex]\displaystyle \int e^-^x \cdot cos2x \ dx = \frac{2 \cdot sin2x-cos2x}{5e^x} + C[/tex].

xKelvin

Answer:

[tex]\displaystyle\int e^{-x}\cos(2x)\, dx=\frac{2\sin(2x)-\cos(2x)}{5e^x}+C[/tex]

Step-by-step explanation:

We would like to integrate the following integral:

[tex]\displaystyle \int e^{-x}\cdot \cos(2x)\, dx[/tex]

Since this is a product of two functions, we can consider using Integration by Parts given by:

[tex]\displaystyle \int u\, dv =uv-\int v\, du[/tex]

So, let’s choose our u and dv. We can choose u base on the following guidelines: LIATE; or, logarithmic, inverse trig., algebraic, trigonometric, and exponential.

Since trigonometric comes before exponential, we will let:

[tex]u=\cos(2x)\text{ and } dv=e^{-x}\, dx[/tex]

By finding the differential of the left and integrating the right, we acquire:

[tex]du=-2\sin(2x)\text{ and } v=-e^{-x}[/tex]

So, our integral becomes:

[tex]\displaystyle \int e^{-x}\cdot \cos(2x)\, dx=(\cos(2x))(-e^{-x})-\int (-e^{-x})(-2\sin(2x))\, dx[/tex]

Simplify:

[tex]\displaystyle \int e^{-x}\cdot \cos(2x)\, dx=-e^{-x}\cos(2x)-2\int e^{-x}\sin(2x)\, dx[/tex]

Since we ended up with another integral of a product of two functions, we can apply integration by parts again. Using the above guidelines, we get that:

[tex]u=\sin(2x)\text{ and } dv=e^{-x}\, dx[/tex]

By finding the differential of the left and integrating the right, we acquire:

[tex]du=2\cos(2x)\, dx\text{ and } v=-e^{-x}[/tex]

This yields:

[tex]\displaystyle \int e^{-x}\cdot \cos(2x)\, dx=-e^{-x}\cos(2x)-2\Big[(\sin(2x))(-e^{-x})-\int (-e^{-x})(2\sin(2x))\, dx\Big][/tex]

Simplify:

[tex]\displaystyle \int e^{-x}\cdot \cos(2x)\, dx=-e^{-x}\cos(2x)-2\Big[-e^{-x}\sin(2x)+2\int e^{-x}\cos(2x)\, dx\Big][/tex]

We can distribute:

[tex]\displaystyle \int e^{-x}\cdot \cos(2x)\, dx=-e^{-x}\cos(2x)+2e^{-x}\sin(2x)-4\int e^{-x}\cos(2x)\, dx[/tex]

The integral on the right is the same as our original integral. So, we can isolate it:

[tex]\displaystyle \Big(\int e^{-x}\cos(2x)\, dx\Big)+4\Big(\int e^{-x}\cos(2x)\, dx)\Big)=-e^{-x}\cos(2x)+2e^{-x}\sin(2x)[/tex]

Combine like integrals:

[tex]\displaystyle 5 \int e^{-x}\cos(2x)\, dx=-e^{-x}\cos(2x)+2e^{-x}\sin(2x)[/tex]

We can factor out an e⁻ˣ from the right:

[tex]\displaystyle 5\int e^{-x}\cos(2x)\, dx=e^{-x}\Big(-\cos(2x)+2\sin(2x)\Big)[/tex]

Dividing both sides by 5 yields:

[tex]\displaystyle \int e^{-x}\cos(2x)\, dx=\frac{e^{-x}}{5}\Big(-\cos(2x)+2\sin(2x)\Big)[/tex]

Rewrite. We of course also need the constant of integration. Therefore, our final answer is:

[tex]\displaystyle\int e^{-x}\cos(2x)\, dx=\frac{2\sin(2x)-\cos(2x)}{5e^x}+C[/tex]