# One of the functions is sin(bx) or cos(bx) or a linear combination of them, the other is an exponential function

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So in those cases we can integrate by parts twice and then reorganise it to get the solution.

Example: $I = \int {{e^x}\sin (x)dx} =$ $\left| {\begin{array}{*{20}{c}} {u = \sin x} \\ {\frac{{dv}}{{dx}} = {e^x}} \\ {\frac{{du}}{{dx}} = \cos x} \\ {v = {e^x}} \end{array}} \right.$ $= {e^x}\sin x - \int {{e^x}\cos x} \,dx =$ $\left| {\begin{array}{*{20}{c}} {u = \cos x} \\ {\frac{{dv}}{{dx}} = {e^x}} \\ {\frac{{du}}{{dx}} = - \sin x} \\ {v = {e^x}} \end{array}} \right.$ $\begin{gathered} = {e^x}\sin x - {e^x}\cos x + \int {{e^x}( - \sin )} \,dx \hfill \\ = {e^x}\sin x - {e^x}\cos x - \int {{e^x}\sin } \,dx \hfill \\ = {e^x}\sin x - {e^x}\cos x - I + {C_1} \hfill \\ \end{gathered}$

So $2I = {e^x}\sin x - {e^x}\cos x + {C_1}$

or $I = \tfrac{1}{2}{e^x}(\sin x - \cos x) + C$ Up a level : Integrals
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