How do you integrate #int e^xcosx# by integration by parts method?

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Answer 1 · 2018-05-04T08:27:20

By integrating by parts twice:

#\frac{e^x(cos(x) + sin(x))}{2}#

We will use the fact that #e^x# remains unchanged when integrated, while the cosine function has the following derivation loop:

#cos(x)-> -\sin(x) -> -\cos(x) -> sin(x) -> cos(x) -> \ldots#

In particular, we're interested in the fact that deriving the cosine function twice means to invert its sign.

So, after a first integration by parts, we have

#int e^xcos(x) = e^x cos(x) - int e^x(-sin(x)) =#
#= e^x cos(x) + int e^xsin(x)#

Now integrate #e^xsin(x)# by parts again: we have

#int e^xsin(x) = e^xsin(x) - \int e^xcos(x)#

Putting all the pieces together, we have

#int e^xcos(x) = e^x cos(x) + e^x sin(x) - \int e^xcos(x)#

As you can see, the (same) integral appears on both sides, this means that we can add it to both sides to get

#2int e^xcos(x) = e^x cos(x) + e^x sin(x)#

and thus solve for it:

#int e^xcos(x) = \frac{e^x cos(x) + e^x sin(x)}{2} = \frac{e^x(cos(x) + sin(x))}{2}#

Answer 2 · 2018-05-04T08:32:37

#int e^x cosx dx = (e^x(cosx+sinx))/2 +C#

As #d(e^x) = e^xdx#, we can integrate by parts int the following way:

#int e^x cosx dx = int cosx d(e^x) #

#int e^x cosx dx = e^xcosx- int e^x d(cosx)#

#int e^x cosx dx = e^xcosx+ int e^x sinx dx#

Integrate by parts again:

#int e^x cosx dx = e^xcosx+ int sinx d(e^x)#

#int e^x cosx dx = e^xcosx+e^xsinx- int e^x d(sinx)#

#int e^x cosx dx = e^xcosx+e^xsinx - int e^x cosx dx#

The integral now appears on both sides of the equation and we can solve for it:

#2int e^x cosx dx = e^xcosx+e^xsinx +C#

#int e^x cosx dx = (e^x(cosx+sinx))/2 +C#