• Oct 31st 2008, 06:34 AM
8614smith
hi
ò(between (pi/6) and (-pi/2)) of (e^x sin(2x))

does anyone know how to solve this definite integral? it seems like everytime i do it i get a different answer,
also does anyone know the MATLAB commands to solve this? as im quite stuck,

any help appreciated,
thanks
• Oct 31st 2008, 07:00 AM
Soroban
Hello, 8614smith!

We're expected to know a certain "by parts" trick . . .

Quote:

$\displaystyle \int^{\frac{\pi}{6}}_{\text{-}\frac{\pi}{2}} e^x\sin2x\,dx$
We have: .$\displaystyle I \;=\;\int e^x\sin2x\,dx$

. . "By parts": .$\displaystyle \begin{array}{ccccccc}u &=& \sin2x & & dv &=& e^x\,dx \\ du &=&2\cos2x\,dx & & v &=& e^x \end{array}$

Then: .$\displaystyle I \;=\;e^x\sin2x - 2\!\int\!e^x\cos2x\,dx$

. . "By parts" again: .$\displaystyle \begin{array}{ccccccc}u &=&\cos2x & & dv &=& e^x\,dx \\ du &=&\text{-}2\sin2x\,dx & & v &=& e^x \end{array}$

And we have: .$\displaystyle I \;=\;e^x\sin2x - 2\bigg[e^x\cos2x + 2\!\int\!e^x\sin2x\,dx\bigg] + C$

. . . . . . . . . . $\displaystyle I \;=\;e^x\sin2x - 2e^x\cos2x - 4\underbrace{\int e^x\sin2x\,dx}_{\text{This is }I} + C$

Hence: .$\displaystyle I \;=\;e^x\sin2x - 2e^x\cos2x - 4I + C$

. . . . . $\displaystyle 5I \;=\;e^x\sin2x - 2e^x\cos2x + C$

. . . . . .$\displaystyle I \;=\;\tfrac{1}{5}e^x(\sin x - 2\cos2x) + C$

Therefore: .$\displaystyle \int e^x\sin2x\,dx \;=\;\tfrac{1}{5}e^x(\sin2x - 2\cos2x) + C\quad\hdots$ ta-DAA!

I'll let you plug in the limits . . .
.
• Nov 1st 2008, 06:53 AM
8614smith
thanks for that i now understand how to do the by parts method, i put the limits in and got -0.58 but i get the answer 1.64 when i put it into MATLAB, a friend said this could be because pi/6 is 30degrees and -pi/2 is -90degrees, do you think i would need to change the limits to 30 and -90? or maybe sin of 30 and -90? to get 1/2 and -1, only the question is supposed to be done without a calculator
thanks
• Nov 1st 2008, 05:52 PM
Soroban
Hello again, 8614smith!

No. do not change the radians to degrees . . .
The radian values are used in the evaluation.

We have: .$\displaystyle \tfrac{1}{5}e^x(\sin2x - 2\cos2x)\bigg|^{\frac{\pi}{6}}_{\text{-}\frac{\pi}{2}}$

. . $\displaystyle = \;\tfrac{1}{5}e^{\frac{\pi}{6}}\bigg[\sin\tfrac{\pi}{3} - 2\cos\tfrac{\pi}{3}\bigg] - \tfrac{1}{5}e^{-\frac{\pi}{2}}\bigg[\sin(\text{-}\pi) - 2\cos(\text{-}\pi)\bigg]$

. . $\displaystyle = \;\tfrac{1}{5}e^{\frac{\pi}{6}}\bigg[\frac{\sqrt{3}}{2} - 2\!\cdot\tfrac{1}{2}\bigg] - \tfrac{1}{5}e^{-\frac{\pi}{2}}\bigg[0 - 2(-1)\bigg]$

. . $\displaystyle = \;\tfrac{1}{5}e^{\frac{\pi}{6}}\left(\frac{\sqrt{3 }}{2}-1\right) - \tfrac{2}{5}e^{-\frac{\pi}{2}}$ . . . etc.