# 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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The important thing here is that the taking the integral (or the derivative) of a sin or cos function will, except possibly for a constant factor be the same as the original function, but with the opposite sign.

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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Next page : We have only one function, but one we can differentiateLast modified: May 20, 2018 @ 11:45