Integral (e^x)cosx dx
Not sure how to start?
Thanks.
$\displaystyle \frac{d}{dx}\left[e^x\cos{x}\right] = e^x\cos{x} - e^x\sin{x}$
So $\displaystyle \int{e^x\cos{x} - e^x\sin{x}\,dx} = e^x\cos{x} + C$
$\displaystyle \int{e^x\cos{x}\,dx} - \int{e^x\sin{x}\,dx} = e^x\cos{x} + C$. Call this equation 1.
Now we need to find $\displaystyle \int{e^x\sin{x}\,dx}$. We use the exact same method.
$\displaystyle \frac{d}{dx}\left[e^x\sin{x}\,dx\right] = e^x\sin{x} + e^x\cos{x}$
So $\displaystyle \int{e^x\sin{x} + e^x\cos{x}\,dx} = e^x\sin{x}$
$\displaystyle \int{e^x\sin{x}\,dx} + \int{e^x\cos{x}\,dx} = e^x\sin{x}$
$\displaystyle \int{e^x\sin{x}\,dx} = e^x\sin{x} - \int{e^x\cos{x}\,dx}$.
Substitute this back into equation 1.
$\displaystyle \int{e^x\cos{x}\,dx} - \left[e^x\sin{x} - \int{e^x\cos{x}\,dx}\right] = e^x\cos{x} + C$
$\displaystyle 2\int{e^x\cos{x}\,dx} - e^x\sin{x} = e^x\cos{x} + C$
$\displaystyle 2\int{e^x\cos{x}\,dx} = e^x\sin{x} + e^x\cos{x} + C$
$\displaystyle \int{e^x\cos{x}\,dx} = \frac{1}{2}e^x(\sin{x} + \cos{x}) + C$.
Note, C can take on any arbitrary constant, but its actual value will change throughout the integration... I've just used the same symbol each time because I'm lazy...