How do you find the integral of #(e^(-x))(cos(2x))#?

Question

HIX Tutor · Accepted Answer

undefined First, let's use integration by parts.
Step 1: Identify u and dv in the formula for integration by parts.
Let u = cos(2x), and dv = e^(-x)dx.
Step 2: Find du and v.
Differentiate u to get du: du = -2sin(2x)dx
Integrate dv to get v: v = -e^(-x)
Step 3: Apply the integration by parts formula ∫udv = uv - ∫vdu.
The integral of (e^(-x))(cos(2x)) is equal to -e^(-x)cos(2x) + ∫2e^(-x)sin(2x)dx.
Step 4: Integrate ∫2e^(-x)sin(2x)dx using integration by parts or another integration technique to find the final solution.

William Abdalla · Answer

#int e^(-x)cos(2x) dx = e^(-x)/5 (-cos(2x) + 2 sin(2x)) + cst#

Use complex numbers. Write #cos(2x) = Re(e^(i 2 x))# So, #int e^(-x) cos(2x) dx = Re int e^(-x)e^(i 2 x) dx = Re int e^((-1+2i)x) dx#

Now it's easy to integrate because you know that if #a ne 0#, #int e^(ax) dx = 1/a e^(ax) + cst#. So :

#int e^(-x)cos(2x) dx = Re(1/(-1+2i) e^((-1+2i)x)) + cst#

After that, you have to simplify #1/(-1+2i) e^((-1+2i)x)# : # e^(-x)/(-1+2i) = e^(-x)(cos(2x) + i sin(2x))/(-1+2i)#

Multiply by #(-1-2i)# everywhere and use the fact that #(a-bi)(a+bi) = a^2+b^2# :

#(cos(2x) + i sin(2x))/(-1+2i) = ((cos(2x) +i sin(2x))(-1-2i))/(1+4)#

Keep only the real part :

#Re((cos(2x) + i sin(2x))/(-1+2i)) = (-cos(2x) + 2 sin(2x))/5#

Finally,

#int e^(-x)cos(2x) dx = e^(-x)/5 (-cos(2x) + 2 sin(2x)) + cst#

Isaiah Allen · Answer

undefined To find the integral of $ e^{-x} \cdot \cos(2x) $, you can use integration by parts. The integration by parts formula is:

$$ \int u \, dv = uv - \int v \, du $$

Let:
$$ u = e^{-x} $$
$$ dv = \cos(2x) \, dx $$

Then:
$$ du = -e^{-x} \, dx $$
$$ v = \frac{1}{2} \sin(2x) $$

Now, plug these into the integration by parts formula:

$$ \int e^{-x} \cdot \cos(2x) \, dx = e^{-x} \cdot \frac{1}{2} \sin(2x) - \int \frac{1}{2} \sin(2x) \cdot (-e^{-x}) \, dx $$

This simplifies to:
$$ = \frac{e^{-x}}{2} \sin(2x) + \frac{1}{2} \int e^{-x} \sin(2x) \, dx $$

Now, to integrate $ \int e^{-x} \sin(2x) \, dx $, you can use integration by parts again:

Let:
$$ u = e^{-x} $$
$$ dv = \sin(2x) \, dx $$

Then:
$$ du = -e^{-x} \, dx $$
$$ v = -\frac{1}{2} \cos(2x) $$

Plug these into the integration by parts formula:

$$ \int e^{-x} \sin(2x) \, dx = -e^{-x} \cdot \frac{1}{2} \cos(2x) - \int \frac{1}{2} \cos(2x) \cdot (-e^{-x}) \, dx $$

$$ = -\frac{e^{-x}}{2} \cos(2x) - \frac{1}{2} \int e^{-x} \cos(2x) \, dx $$

Now, let's bring the $ \int e^{-x} \cos(2x) \, dx $ term to the other side:

$$ \frac{3}{2} \int e^{-x} \cos(2x) \, dx = \frac{e^{-x}}{2} \sin(2x) - \frac{e^{-x}}{2} \cos(2x) $$

$$ \int e^{-x} \cos(2x) \, dx = \frac{2}{3} \left( \frac{e^{-x}}{2} \sin(2x) - \frac{e^{-x}}{2} \cos(2x) \right) + C $$

$$ = \frac{e^{-x}}{3} \sin(2x) - \frac{e^{-x}}{3} \cos(2x) + C $$

So, the integral of $ e^{-x} \cos(2x) $ is $ \frac{e^{-x}}{3} \sin(2x) - \frac{e^{-x}}{3} \cos(2x) + C $, where $ C $ is the constant of integration.