Question

Evaluate :$\int e^{-3x}{\cos }^{3}xdx$

Answer 1

Given : $\int e^{-3x}{\cos }^{3}xdx$
We know, $\cos 3x=4{\cos }^{3}x-3\cos x$

$\int e^{-3x}{\cos }^{3}xdx$
$=\int e^{-3x}\left[\frac{\cos 3x+3\cos x}{4}\right]dx$
$=\frac{1}{4}\int e^{-3x}\cos 3xdx+\frac{3}{4}\int e^{-3x}\cos xdx$

Now, finding the integration of
$I=\int e^{ax}\cos bxdx$

Using integration by parts, we get
$\Rightarrow I=\cos bx\int e^{ax}dx-\int \left[\frac{d}{dx}\right(\cos bx)\int e^{ax}dx]dx$

$\Rightarrow I=\cos bx\left(\frac{e^{ax}}{a}\right)+\frac{b}{a}\int (\sin bx\cdot e^{ax})dx$

Using integration by parts, we get

$\Rightarrow I=\frac{1}{a}{e}^{ax}\cos bx+\frac{b}{a}[\sin bx\int e^{ax}dx-\int \left[\frac{d}{dx}\right(\sin bx)\int e^{ax}dx]dx]$

$\Rightarrow I=\frac{1}{a}{e}^{ax}\cos bx+\frac{b}{a}[\sin bx\left(\frac{e^{ax}}{a}\right)-\frac{b}{a}\int (\cos bx\cdot e^{ax}\left)dx\right]$

$\Rightarrow I=\frac{1}{a}{e}^{ax}\cos bx+\frac{b}{a^2}{e}^{ax}\sin bx-\frac{b^2}{a^2}\int (\cos bx\cdot e^{ax})dx$

$\Rightarrow I=\frac{1}{a^2}{e}^{ax}(a\cos bx+b\sin bx)-\frac{b^2}{a^2}I$

$\Rightarrow (1+\frac{b^2}{a^2})I=\frac{1}{a^2}{e}^{ax}(a\cos bx+b\sin bx)$

$\therefore \int e^{ax}\cos bxdx$

$=\frac{1}{a^2+b^2}{e}^{ax}(a\cos bx+b\sin bx)$

Now, equation (1) becomes
$\frac{1}{4}\int e^{-3x}\cos 3xdx+\frac{3}{4}\int e^{-3x}\cos xdx$

$=\frac{1}{4}\left[\frac{1}{(-3{)}^2+(3{)}^2}{e}^{-3x}\right(-3\cos 3x+3\sin 3x\left)\right]+\frac{3}{4}\left[\frac{1}{(-3{)}^2+(1{)}^2}{e}^{-3x}\right(-3\cos x+\sin x\left)\right]+c$

$=\frac{3}{4\times 18}{e}^{-3x}(\sin 3x-\cos 3x)+\frac{3}{4\times 10}{e}^{-3x}(\sin x-3\cos x)+c$

$=\frac{e^{-3x}}{24}(\sin 3x-\cos 3x)+\frac{3e^{-3x}}{40}(\sin x-3\cos x)+c$
where, $c$ is constant of integration

Hence, the integration is
$\frac{e^{-3x}}{24}(\sin 3x-\cos 3x)+\frac{3e^{-3x}}{40}(\sin x-3\cos x)$