So i wany to integrate the function $e^x.cos(x)$ so i decided to do it like this way $e^x.e^{ix}$ and then only taking the real part from it so i get $\frac{(1-i)}{2}(cos(x)+sin(x))$ and then taking the real part . now i wanted to verify my result so i entered $e^x.e^{i \cdot x}$ in WA and it said no results found in elementary functions but when i wrote it as $e^{x(1+i)}$ it did calculate what is the reason behind it . the complex numbers do obey laws of indices then where did i go wrong. Also tell me whether theres a more good way to do $$\int e^x.\cos(x)$$
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See http://math.stackexchange.com/questions/20952/integration-by-parts-int-eax-cosbx-dx – lab bhattacharjee Mar 15 '16 at 06:21
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No but integration by parts makes me go round and round can you tell me how to do the next part – Archis Welankar Mar 15 '16 at 06:24
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3Let $I$ be your integral. Apply Parts twice. Realize you have $I$ plus some other stuff. "Going round and round" means you're passing by the solution on every revolution. – Eric Towers Mar 15 '16 at 06:26
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You used the proper trick for this kind od integrals. Considering $$I=\int e^x \cos(x)\,dx \qquad J=\int e^x \sin(x)\,dx \implies I+iJ=\int e^x e^{ix}\,dx=\int e^{(1+i)x}\,dx$$ $$I+iJ=\frac{e^{(1+i)x}}{1+i}=\frac 12(1-i)e^x e^{ix}=\frac 12(1-i)e^x (\cos(x)+i \sin(x))$$ Expanding and grouping real and imaginary terms then leads to $$I+iJ=\frac 12e^x (\cos(x)+\sin(x))+i\frac 12e^x(\sin(x)-\cos(x))$$ then $I$ and $J$.
Claude Leibovici
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