Let INT = integral symbol
INT( e^x)(cosx) dx
What is u and dv here to get me started?
use ILATE LAW
if u=first fn(higher precedence) which here is of cos(x)
v=second fn IS e^(x)
now
I=COS(X)INTEGRATE(E^(X))-INTEGRATE(DIFF(COS(X)INTEGRATE(E^(X)))
I=COS(X)E^(X)+INTEGRATE(SIN(X)E^(X))
NOW APPLY INTEGRATION OF PARTS ONE MORE TIME
I=COS(X)E^(X)+E^(X)SIN(X)-I WHERE I=INTEGRATE(E^(X)COS(X)) [UNDERSTAND IN TERMS OF RECURSION]
SOLVING I U GET
I=E^(X)(COS(X)+SIN(X))/2+CONSTANT
Either way will work:
a) Let $\displaystyle u= e^x$ and $\displaystyle dv= cos(x) dx$. Then $\displaystyle du= e^x dx$ and $\displaystyle v= sin(x)$ so we have $\displaystyle \int e^x cos(x)dx= uv- \int v du= (e^x sin(x))- \int e^x sin(x) dx$. To integrate $\displaystyle \int e^x sin(x)$, let $\displaystyle u= e^x$ and $\displaystyle dv= sin(x)dx$. Then $\displaystyle du= e^x dx$ and $\displaystyle v= -cos(x)$ so we have $\displaystyle \int e^x sin(x)dx= uv- \int vdu= -e^x cos(x)+ \int e^x cos(x)dx$.
Putting those together,
$\displaystyle \int e^x cos(x)dx= e^x sin(x)- (-e^x cos(x)+ e^x cos(x)dx)= e^x(sin(x)+ cos(x))- \int e^x cos(x)dx$
and then $\displaystyle 2\int e^x cos(x)dx= e^x(sin(x)+ cos(x))$,
$\displaystyle \int e^x cos(x)dx= \frac{1}{2}e^{x}(sin(x)+ cos(x))+ C$.
Now, you try it letting $\displaystyle u= cos(x)$, $\displaystyle dv= e^x dx$.
(By the way- when you are not sure which of two functions to take as "u" and which to take as "dv", try both!)
You seem to be missing the point- whatever excuse (reason) you have for not using Latex, prasum, or any other person here, may have the same reason. Who are to ask that other people do what you, for whatever reason, do not do?