# How to calculate the integral?

## ${\int}_{0}^{\frac{\pi}{3}}$$\sin x \cdot {e}^{- 3 x}$dx

May 5, 2018

$I = \frac{1}{10} \left[1 - \frac{3 \sqrt{3} + 1}{2 {e}^{\pi}}\right]$
$I \approx - 2.214069$

#### Explanation:

Note:

color(red)(inte^(ax)sinbxdx=(e^(ax))/(a^2+b^2)(asinbx-bcosbx)+c

Take,$a = - 3 \mathmr{and} b = 1.$

inte^(-3x)sin(1x)dx=(e^(-3x))/((-3)^2+1^2)(-3sinx- 1cos(1x))+c

color(blue)(inte^(-3x)sinxdx=(e^(-3x))/(10)(-3sinx-cosx)+c...to (1)

Here,

$I = {\int}_{0}^{\frac{\pi}{3}} \sin x \cdot {e}^{- 3 x} \mathrm{dx}$

$I = {\int}_{0}^{\frac{\pi}{3}} \left({e}^{- 3 x}\right) \sin x \mathrm{dx}$

Using $\left(1\right)$,we get

$I = {\left[\frac{{e}^{- 3 x}}{10} \left(- 3 \sin x - \cos x\right)\right]}_{0}^{\frac{\pi}{3}}$

I=[(e^(-3xxpi/3))/10(-3sin(pi/3)-cos(pi/3)]-[e^0/10(-3(0)-1)]

$I = \left[{e}^{-} \frac{\pi}{10} \left(- 3 \times \frac{\sqrt{3}}{2} - \frac{1}{2}\right)\right] - \left[\frac{1}{10} \left(- 1\right)\right]$

$I = {e}^{-} \frac{\pi}{10} \left(- \frac{3 \sqrt{3} + 1}{2}\right) + \frac{1}{10}$

$I = \frac{1}{10} \left[1 - \frac{3 \sqrt{3} + 1}{2 {e}^{\pi}}\right]$

$I \approx \frac{1}{10} \left[1 - 23.14069\right]$

$I \approx \frac{1}{10} \left[- 22.14069\right]$

$I \approx - 2.214069$