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

2 Answers
Dec 18, 2016

Answer:

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

Explanation:

#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#

Notice that we have the integral we started out with on both sides of the equation now--that is, we can 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#

Jul 30, 2017

Answer:

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

Explanation:

Another approach when dealing with integrals of the form:

# I_1 = int \ e^(ax)sin(omega x) \ \ #, or # \ \ I_2 = int \ e^(ax)cos(omega x) #

Is to use some intuition to determine the form of the solution.

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 doesn't help much until we apply Integration By Parts a second time, giving:

# 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)) #

Knowing this, we can start with an assumption of the integral result and differentiate it to see if it works, thus:

Assume a solution of the form:

# 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 #

Solving these simultaneous equations we get

# A=2/5, \ B=-1/5 #

Thus we have:

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

Hence as this is an antiderivative of our initial integral, and we have:

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