# Solve Integral Analytically

• February 19th 2009, 05:22 PM
Cohsty243
Solve Integral Analytically
Integral from -2 to 3 of e^(2x)cos(3x)dx

I think I am suppose to solve for the unknown integral using Integration by parts formula however I have tried multiplies times but no luck lol.
• February 20th 2009, 04:58 AM
DeMath
Quote:

Originally Posted by Cohsty243
Integral from -2 to 3 of e^(2x)cos(3x)dx

I think I am suppose to solve for the unknown integral using Integration by parts formula however I have tried multiplies times but no luck lol.

$I = \int {{e^{2x}}\cos \left( {3x} \right)dx} = \frac{{{e^{2x}}}}{3}\sin \left( {3x} \right) - \frac{2}{3}\int {{e^{2x}}\sin \left( {3x} \right)dx} =$

$= \frac{{{e^{2x}}}}{3}\sin \left( {3x} \right) + \frac{{2{e^{2x}}}}{9}\cos \left( {3x} \right) - \frac{4}{9}\int {{e^{2x}}\cos \left( {3x} \right)dx} .$

$I = \frac{{{e^{2x}}}}{3}\sin \left( {3x} \right) + \frac{{2{e^{2x}}}}{9}\cos \left( {3x} \right) - \frac{4}{9}I \Leftrightarrow$

$\Leftrightarrow I + \frac{4}{9}I = \frac{{{e^{2x}}}}{3}\sin \left( {3x} \right) + \frac{{2{e^{2x}}}}{9}\cos \left( {3x} \right) \Leftrightarrow$

$\Leftrightarrow \frac{{13}}{9}I = \frac{{{e^{2x}}}}{9}\left[ {3\sin \left( {3x} \right) + 2\cos \left( {3x} \right)} \right] \Leftrightarrow$

$\Leftrightarrow \boxed{I = \frac{{{e^{2x}}}}{{13}}\left[ {3\sin \left( {3x} \right) + 2\cos \left( {3x} \right)} \right] + C}$

Please note that the function $f\left( x \right) = {e^{2x}}\cos \left( {3x} \right)$ has five roots

${x_1} = - \frac{\pi }{2};{\text{ }}{x_2} = - \frac{\pi }{6};{\text{ }}{x_3} = \frac{\pi }{6};{\text{ }}{x_4} = \frac{\pi }{2};{\text{ }}{x_5} = \frac{{5\pi }}{6}$

on the interval $- 2 \leqslant x \leqslant 3$, ie, this function $6$ times changes its sign in this interval. I hope you understand why this is necessary.

If you calculate everything correctly, then your response should be approximately equal to $117.086$.
• February 21st 2009, 11:50 AM
Cohsty243
Thank you (Clapping)