How do I find the integral #int(cos(x)/e^x)dx# ?

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Answer 1 · 2014-09-21T14:53:25

#int{cosx}/e^xdx={e^{-x}}/2(sinx-cosx)+C#

Let us look at some details.

Let

#I=int{cos x}/{e^x}dx#

by rewritng #1/e^x# as #e^{-x}#,

#=int e^{-x}cos x dx#

by Integration by Parts

Let #u=e^{-x}# and #dv=cosx dx#
#Rightarrow du=-e^{-x}dx# and #v=sin x#

#=e^{-x}sinx+int e^{-x}sinx dx#

by another Integration by Parts,

Let #u=e^{-x}# and #dv=sinx dx#
#Rightarrow du=-e^{-x}dx# and #v=-cosx#

#=e^{-x}sinx-e^{-x}cosx-I#

So, we have

#I=e^{-x}(sinx-cosx)-I#

by adding #I#,

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

by dividing by 2,

#I={e^{-x}}/2(sinx-cosx)+C#