How do you find the antiderivative of #e^(2x)(sin x)#?
2 Answers
Explanation:
#I=inte^(2x)sinxdx#
We should try integration by parts. Typically when assigning values of
In fact, we see it doesn't really matter which we choose for
#{(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
#{(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
#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#
# 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 = 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 = 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
# dy/dx = e^(2x)(Acosx-Bsinx) + 2e^(2x)(Asinx + Bcosx)#
# " " = e^(2x)((A+2B)cosx+(2A-B)sinx) #
We want
# 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#