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hytuoc
Sep24-04, 07:35 PM
Some one plz show me how to do this problem below
Integral of e^x cos(x) dx
how to I integrate that?
thanks

Pyrrhus
Sep24-04, 07:40 PM
Try Integration by parts

\int u dv = uv -\int v du

try using u = e^x and dv = \cos (x) dx

Hurkyl
Sep24-04, 07:44 PM
Your book will have this example (or one nearly identical to it) perfored step by step.

Phymath
Sep24-04, 07:47 PM
u = e^x \ dv = cos(x) dx
du = e^x dx \ v = sin(x)

\int e^x cos(x) dx = e^x sin(x) - \int sin(x) e^x dx

u = e^x \ dv = sin(x) dx
du = e^x dx \ v = -cos(x) dx
\int e^x cos(x) dx = e^x sin(x) - (-e^x cos(x) + \int e^x cos(x) dx)
= e^x sin(x) + e^x cos(x) - \int e^x cos(x) dx
\int e^x cos(x) dx + \int e^x cos(x) = e^x (sin(x) + cos(x))
2 \int e^x cos(x) dx = e^x(sin(x) + cos(x))
\int e^x cos(x) dx = 1/2 e^x (sin(x) + cos(x)) there you go intergration by parts follow
u v - \int v du

hytuoc
Sep24-04, 08:23 PM
thanks so much