Integrate e^x cos(x)dx, how to do this?

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To integrate e^x cos(x) dx, use integration by parts with u = e^x and dv = cos(x) dx. This leads to the equation ∫ e^x cos(x) dx = e^x sin(x) - ∫ e^x sin(x) dx. Applying integration by parts again on the remaining integral results in the equation ∫ e^x cos(x) dx + ∫ e^x cos(x) dx = e^x (sin(x) + cos(x)). Simplifying gives 2 ∫ e^x cos(x) dx = e^x (sin(x) + cos(x)), leading to the final result of ∫ e^x cos(x) dx = 1/2 e^x (sin(x) + cos(x)). This method effectively demonstrates the application of integration by parts.
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Some one please show me how to do this problem below
Integral of e^x cos(x) dx
how to I integrate that?
thanks
 
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Try Integration by parts

\int u dv = uv -\int v du

try using u = e^x and dv = \cos (x) dx
 
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Your book will have this example (or one nearly identical to it) perfored step by step.
 
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
 
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thanks so much
 
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