Question

Evaluate ∫e to the power x (sin(4x))dx

Answer 1

Step-by-step explanation:

We have,

$\tt{ Let \: \: I = \int {e}^{x} \sin(4x) dx }$

$\sf{\implies I = {e}^{x} \int sin(4x) dx - \int \bigg \{ \frac{d}{dx} ( {e}^{x} ) \int \: sin(4x) \: dx \bigg \}dx }$

$\sf{\implies I = - {e}^{x} \cdot\frac{ cos(4x) }{4} + \int {e}^{x} \cdot\frac{cos(4x)}{4} dx }$

$\sf{\implies I = - \frac{1}{4} {e}^{x} cos(4x) + \frac{1}{4} \int {e}^{x} \: cos(4x)dx }$

$\sf{\implies I = - \frac{1}{4} {e}^{x} cos(4x) + \frac{1}{4} \: {e}^{x} \int \: cos(4x)dx - \frac{1}{4} \int \bigg \{ \frac{d}{dx} ( {e}^{x}) \int \: cos(4x) \: dx \bigg \}dx }$

$\sf{\implies I = - \frac{1}{4} {e}^{x} cos(4x) + \frac{1}{4} \: {e}^{x} \cdot \: \frac{sin(4x)}{4}- \frac{1}{4} \int {e}^{x} \cdot\: \frac{ sin(4x)}{4} dx }$

$\sf{\implies I = - \frac{1}{4} {e}^{x} cos(4x) + \frac{1}{16} \: {e}^{x} sin(4x)- \frac{1}{16} \int {e}^{x} sin(4x) dx }$

$\sf{\implies I = - \frac{1}{4} {e}^{x} cos(4x) + \frac{1}{16} \: {e}^{x} sin(4x)- \frac{1}{16} \: I }$

$\sf{\implies I + \frac{1}{16} \: I = - \frac{1}{4} {e}^{x} cos(4x) + \frac{1}{16} \: {e}^{x} sin(4x) }$

$\sf{\implies \frac{17}{16} \: I = - \frac{1}{4} {e}^{x} cos(4x) + \frac{1}{16} \: {e}^{x} sin(4x) }$

$\sf{\implies 17 \: I = - 4 {e}^{x} cos(4x) + \: {e}^{x} sin(4x) }$

$\sf{\implies \: I = \frac{1}{17} \{ {e}^{x} sin(4x) } - 4 {e}^{x} cos(4x) \}$

$\sf{\implies \: I = \frac{ {e}^{x} }{17} \{ sin(4x) } - 4 cos(4x) \}$

$\sf{\implies \: I = \frac{ {e}^{x} }{ \sqrt{ 17}} \bigg\{ \frac{1}{ \sqrt{17} } sin(4x) - \frac{4}{ \sqrt{17} } cos(4x) \bigg\}}$

Let $\rm{tan(\alpha)=4}$

So,

$\rm{cos(\alpha)=\dfrac{1}{\sqrt{17}}\,\,\,\,\&\,\,\,\, sin(\alpha)=\dfrac{4}{\sqrt{17}}}$

So,

$\sf{\implies \: I = \frac{ {e}^{x} }{ \sqrt{ 17}} \{ sin(4x) cos( \alpha) - cos(4x)sin( \alpha) \}}$

$\sf{\implies \: I = \frac{ {e}^{x} }{ \sqrt{ 17}} \: sin(4x - \alpha) }$

$\sf{\implies \: I = \frac{ {e}^{x} }{ \sqrt{ 17}} \: sin(4x - tan^{ - 1} (4)) }$