Question

Evaluate the following:
$\int e^{-3x}co{s}^{3}xdx$

Answer 1

Given $I=\int e^{-3x}{cos}^{3}xdx$

$\therefore I=\int e^{-3x}\left(\frac{cos3x+3cosx}{4}\right)dx$

$\therefore I=\frac{1}{4}\int e^{-3x}(cos3x+3cosx)dx$

$\therefore I=\frac{1}{4}[\int e^{-3x}cos3xdx+3\int e^{-3x}cosxdx]$ (1)

Let $I_1=\int e^{-3x}cos3xdx$

$\therefore I_1=cos3x\int e^{-3x}dx-\int [\frac{d}{dx}\left(cos3x\right)\int e^{-3x}dx]dx$

$\therefore I_1=cos3x\frac{e^{-3x}}{-3}-\int [-3sin3x\left(\frac{e^{-3x}}{-3}\right)]dx$

$\therefore I_1=\frac{e^{-3x}cos3x}{-3}-\int \left[e^{-3x}sin3x\right]dx$

$\therefore I_1=\frac{e^{-3x}cos3x}{-3}-\{sin3x\int e^{-3x}dx-\int \frac{d}{dx}\left(sin3x\right)\int e^{-3x}dx\}$

$\therefore I_1=\frac{e^{-3x}cos3x}{-3}-\{sin3x\frac{e^{-3x}}{-3}-\int 3cos3x\frac{e^{-3x}}{-3}dx\}$

$\therefore I_1=\frac{e^{-3x}cos3x}{-3}-\{\frac{e^{-3x}sin3x}{-3}+\int e^{-3x}cos3xdx\}$

$\therefore I_1=\frac{e^{-3x}cos3x}{-3}-\{\frac{e^{-3x}sin3x}{-3}+I_1\}$

$\therefore I_1=\frac{e^{-3x}cos3x}{-3}+\frac{e^{-3x}sin3x}{3}-I_1$

$\therefore 2I_1=\frac{e^{-3x}}{3}[sin3x-cos3x]$

$\therefore I_1=\frac{e^{-3x}}{6}[sin3x-cos3x]$

Now, let $I_2=\int e^{-3x}cosxdx$

$\therefore I_2=cosx\int e^{-3x}dx-\int [\frac{d}{dx}\left(cosx\right)\int e^{-3x}dx]dx$

$\therefore I_2=cosx\frac{e^{-3x}}{-3}-\int [-sinx\left(\frac{e^{-3x}}{-3}\right)]dx$

$\therefore I_2=\frac{e^{-3x}cosx}{-3}-\frac{1}{3}\int e^{-3x}sinxdx$

$\therefore I_2=\frac{e^{-3x}cosx}{-3}-\frac{1}{3}\{sinx\int e^{-3x}dx-\int [\frac{d}{dx}\left(sinx\right)\int e^{-3x}dx]dx\}$

$\therefore I_2=\frac{e^{-3x}cosx}{-3}-\frac{1}{3}\{sinx\frac{e^{-3x}}{-3}-\int \left[cosx\left(\frac{e^{-3x}}{-3}\right)\right]dx\}$

$\therefore I_2=\frac{e^{-3x}cosx}{-3}-\frac{1}{3}\{sinx\frac{e^{-3x}}{-3}+\frac{1}{3}\int e^{-3x}cosxdx\}$

$\therefore I_2=\frac{e^{-3x}cosx}{-3}+\frac{1}{9}{e}^{-3x}sinx-\frac{1}{9}{I}_2$

$\therefore I_2+\frac{1}{9}{I}_2=\frac{e^{-3x}cosx}{-3}+\frac{1}{9}{e}^{-3x}sinx$

$\therefore \frac{10}{9}{I}_2=\frac{e^{-3x}cosx}{-3}+\frac{1}{9}{e}^{-3x}sinx$

$\therefore I_2=\frac{9}{10}[\frac{e^{-3x}cosx}{-3}+\frac{1}{9}{e}^{-3x}sinx]$

$\therefore I_2=\frac{-3}{10}{e}^{-3x}cosx+\frac{1}{10}{e}^{-3x}sinx$

$\therefore I_2=\frac{e^{-3x}}{10}[sinx-3cosx]$

Thus, from (1),

$I=\frac{1}{4}[\frac{e^{-3x}}{6}[sin3x-cos3x]+3\frac{e^{-3x}}{10}[sinx-3cosx]]$

$\therefore I=\frac{e^{-3x}}{24}[sin3x-cos3x]+\frac{3e^{-3x}}{40}[sinx-3cosx]$