Calculus - Very simple, yet very often ignored method

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I=âˆ«e^xcosx = real part of âˆ«e^xe^ix dx
= real part of âˆ«e^(1+i)x
=real part of e^(1+i)x/(1+i)+C₁+iC₂....

62

yup that is what i was looking for :)

62

For Sky.. to make this integration more clear to you...

e^xcosx = Re[e^x(cosx+i sinx)]

Re[e^x(cosx+i sinx)]= Re[e^x.e^ix] =Re[e^x+ix]

e^xcosx=Re[e^x+ix]
âˆ«e^xcosx=âˆ«Re[e^x+ix]
=Re[âˆ«e^x+ix]

=Re[âˆ«e^(1+i)x]

= Re[(e^x+ix)/(1+i)]

Now we just need to calcualte the real part of this given number!

1

thank you!
got it...
clear!!!

Very simple, yet very often ignored method