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

Up a level : Integrals
Previous page : One function is in the form x to the power of n and the other can be integrated at least n times
Next page : We have only one function, but one we can differentiate

The important thing here is that 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:

$I = \int {{e^x}\sin (x)dx} =$

$\left| {\begin{array}{*{20}{c}} {u = \sin x} \\ {\frac{{dv}}{{dx}} = {e^x}} \\ {\frac{{du}}{{dx}} = \cos x} \\ {v = {e^x}} \end{array}} \right.$

$= {e^x}\sin x - \int {{e^x}\cos x} \,dx =$

$\left| {\begin{array}{*{20}{c}} {u = \cos x} \\ {\frac{{dv}}{{dx}} = {e^x}} \\ {\frac{{du}}{{dx}} = - \sin x} \\ {v = {e^x}} \end{array}} \right.$

$\begin{gathered} = {e^x}\sin x - {e^x}\cos x + \int {{e^x}( - \sin )} \,dx \hfill \\ = {e^x}\sin x - {e^x}\cos x - \int {{e^x}\sin } \,dx \hfill \\ = {e^x}\sin x - {e^x}\cos x - I + {C_1} \hfill \\ \end{gathered}$

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

$I = \tfrac{1}{2}{e^x}(\sin x - \cos x) + C$

Last modified: Dec 28, 2023 @ 14:33