How do you find the antiderivative of #e^(2x)(sin x)#?

Question

Christopher Allers · Accepted Answer

#inte^(2x)sinxdx=1/5e^(2x)(2sinx-cosx)+C#

#I=inte^(2x)sinxdx#

We should try integration by parts. Typically when assigning values of #u# and #dv#, we want to choose a function for #u# that will get simpler as we differentiate it. However, we see that #e^(2x)# will stay being an exponential function and #sinx# will bounce back and forth through trigonometric functions.

In fact, we see it doesn't really matter which we choose for #u# and which for #dv#. On a whim, let:

#{(u=e^(2x),=>,du=2e^(2x)dx),(dv=sinxdx,=>,v=-cosx):}#

Then:

#I=uv-intvdu#

#I=-e^(2x)cosx-int(-cosx)(2e^(2x)dx)#

#I=-e^(2x)cosx+2inte^(2x)cosxdx#

Perform integration by parts once more. Again choose #e^(2x)# as #u#.

#{(u=e^(2x),=>,du=2e^(2x)dx),(dv=cosxdx,=>,v=sinx):}#

#I=-e^(2x)cosx+2[uv-intvdu]#

#I=-e^(2x)cosx+2uv-2intvdu#

#I=-e^(2x)cosx+2e^(2x)sinx-2intsinx(2e^(2x)dx)#

#I=-e^(2x)cosx+2e^(2x)sinx-4inte^(2x)sinxdx#

Observe that both sides of the equation now contain the integral we began with, allowing us to write:

#I=-e^(2x)cosx+2e^(2x)sinx-4I#

Solve for #I# treating the entire integral like we would any other variable:

#5I=-e^(2x)cosx+2e^(2x)sinx#

#5I=e^(2x)(2sinx-cosx)#

#I=1/5e^(2x)(2sinx-cosx)#

#inte^(2x)sinxdx=1/5e^(2x)(2sinx-cosx)*C#

Scarlett Allison · Answer

# int \ e^(2x)sin(x) \ dx = e^(2x)(2/5sinx- 1/5cosx) + c#

An alternative method for handling form integrals is as follows: # I_1 = int \ e^(ax)sin(omega x) \ \ #, or # \ \ I_2 = int \ e^(ax)cos(omega x) # is to determine the solution's form by applying some intuition. Irrespective of the trig function, the results are analogous, so wlog let us consider only #I_1#. If we use Integration by parts we find that we can decompose #I_1# as follows: # I_1 = A_1e^(ax)cos(omega x) + A_2 \ int \ e^(ax)cos(omega x) # which is not very helpful until we use Integration By Parts twice, providing: # I_1 = A_4e^(ax)sin(omega x) + A_5e^(ax)cos(omega x) + A_6 \ int \ e^(ax)sin(omega x) # Or: # I_1 = A_4e^(ax)sin(omega x) + A_5e^(ax)cos(omega x) + A_6 I_1 # Which is now an algebraic equation which can be solved for #I_1# # I_1 = e^(ax)(A_7sin(omega x) + A_8cos(omega x)) # With this knowledge, we can begin by making the following assumption about the integral result and then differentiate it to see if it works: Assume the following solution: # y= e^(2x)(Asinx + Bcosx)# Differentiating wrt #x# and applying the product rule we get: # dy/dx = e^(2x)(Acosx-Bsinx) + 2e^(2x)(Asinx + Bcosx)# # " " = e^(2x)((A+2B)cosx+(2A-B)sinx) # We want #dy/dx = e^(2x)sinx#, so equating coefficients of sine and cosine we get: # cosx: \ A+2B=0 # # sinx: \ 2A-B=1 # When we solve these concurrent equations, we obtain # A=2/5, \ B=-1/5 # Consequently, we have: # y= e^(2x)(2/5sinx- 1/5cosx)# Since this is our initial integral's antiderivative, we have: # int \ e^(2x)sin(x) \ dx = e^(2x)(2/5sinx- 1/5cosx) + c#

Isaiah Allen · Answer

undefined To find the antiderivative of $ e^{2x} \cdot \sin(x) $, you can use integration by parts. Let $ u = \sin(x) $ and $ dv = e^{2x} \, dx $. Then, $ du = \cos(x) \, dx $ and $ v = \frac{1}{2} e^{2x} $. Applying the integration by parts formula:

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

Now, we need to integrate $ \frac{1}{2} e^{2x} \cdot \cos(x) $. This can be done by integration by parts again. Let $ u = \cos(x) $ and $ dv = \frac{1}{2} e^{2x} \, dx $. Then, $ du = -\sin(x) \, dx $ and $ v = \frac{1}{4} e^{2x} $. Applying the integration by parts formula:

$$ \int e^{2x} \cdot \sin(x) \, dx = -\frac{1}{2} e^{2x} \cdot \cos(x) + \frac{1}{2} \left( \frac{1}{4} e^{2x} \cdot \cos(x) + \frac{1}{4} \int e^{2x} \cdot \sin(x) \, dx \right) $$

Now, we have:

$$ \int e^{2x} \cdot \sin(x) \, dx = -\frac{1}{2} e^{2x} \cdot \cos(x) + \frac{1}{8} e^{2x} \cdot \cos(x) + \frac{1}{8} \int e^{2x} \cdot \sin(x) \, dx $$

Solving for the integral, we get:

$$ \frac{7}{8} \int e^{2x} \cdot \sin(x) \, dx = \frac{1}{2} e^{2x} \cdot \cos(x) $$

$$ \int e^{2x} \cdot \sin(x) \, dx = \frac{2}{7} e^{2x} \cdot \cos(x) + C $$

Where $ C $ is the constant of integration.