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$$\int_0^5 e^{-2t}\sin(t) \,\mathrm{d}t$$

I know I should be able to integrate this by parts but I can't seem to get the parts I choose to make the result any easier to integrate.

Olivier Oloa
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Chuck S.
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    Doing it by parts once gets you an integral with cos instead of sin...so if you integrate by parts again... – lulu Apr 24 '16 at 18:34
  • Denote the original integral $I$ and save time! –  Apr 24 '16 at 18:35
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    Another way to achieve this is to evaluate $\int_0^{5} e^{-2t+it}dt$ and take the imaginary part. – Evariste Apr 24 '16 at 18:45

5 Answers5

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In fact, you're not meant to find something easier to integrate.

After integrating by parts twice (keep integrating $e^{-2t}$ and deriving the trigonometric part), you'll end up with something like $$\int_0^5 e^{-2t}\sin t\,dt=\text{something }+k\int_0^5e^{-2t}\sin t\,dt$$ with $k\ne 1$.

At that point, you just need to do this $$(1-k)\int_0^5 e^{-2t}\sin t\,dt=\text{something }\\\int_0^5 e^{-2t}\sin t\,dt=\frac{\text{something}}{1-k}$$

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    Nice. Please keep the "something" in there...It's a good exercise+1 – imranfat Apr 24 '16 at 18:35
  • Right, right. I gave up too soon in integration by parts. Believe it or not, in another math book I have I looked at example, integral of e^x sin x dx and saw the trick of integrating by parts twice by integrating the integral left over from the first integration, again. I saw the error of my ways and integrated my original problem, above, successfully. Thanks for being patient. I am 70 and took this math 50 years ago and am reviewing. This was all part of a proof on evaluation of integral 0 to inf sin x/x in my old book by Watson Fulks. He can get a bit deep, for me. – Chuck S. Apr 24 '16 at 19:09
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This integral is the imaginary part of \begin{align*}\int_0^5\mathrm e^{(-2+\mathrm i)t}\,\mathrm d \mkern1mu t&=\frac1{-2+\mathrm i}\mathrm e^{(-2+\mathrm i)t}\biggr\rvert_0^5=-\frac{2+ \mathrm i}{5}\bigl(\mathrm e^{-10+5\mathrm i}-1)\\ &=-\frac15\bigl(2(\mathrm e^{-10}\cos 5-1)-\mathrm e^{-10}\sin 5+\mathrm i(\mathrm e^{-10}\cos 5-1+2\mathrm e^{-10}\sin 5)\bigr), \end{align*} so the answer is $$\frac{1-\mathrm e^{-10}(\cos 5+2\sin 5)}5.$$

Bernard
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By indeterminate coefficients:

By educated guess, you try a solution of the form

$$e^{-2t}(a\cos(t)+b\sin(t)).$$

Deriving, you get

$$e^{-2t}(-2a\cos(t)-2b\sin(t)-a\sin(t)+b\cos(t)).$$

The unknown coefficients are obtained by solving

$$-2a+b=0,\\-2b-a=1,$$ giving

$$a=-\frac15,b=-\frac25.$$

The definite integral easily follows.

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Hint. Integrating by parts twice gives $$ \begin{align} \int_0^5e^{-2t}\sin t\:dt&=\left. -\frac12e^{-2t}\sin t\right|_0^5+\frac{1}{2}\int_0^5e^{-2t}\cos t\:dt\ \end{align} $$ and $$ \begin{align} \int_0^5e^{-2t}\cos t\:dt&=\left. -\frac12e^{-2t}\cos t\right|_0^5-\frac{1}{2}\int_0^5e^{-2t}\sin t\:dt\ \end{align} $$

Can you take it from here?

Olivier Oloa
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  • I select this as the best answer for ME, emphasizing me. The other answers are all fine, but some were not quite as readily understandable to me, as I am reviewing math from 50 years ago and am a bit rusty at present. – Chuck S. May 07 '16 at 19:43
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Since we can do integrals similarly to $\mathbb{R}\to\mathbb{C}$ as well: $$ \sin x=\frac{e^{xi}-e^{-xi}}{2i}\qquad e^{a+bi}=e^a\left(\cos b+i\sin b\right) $$ So with a more direct approach: $$ \int_0^5e^{-2t}\sin t\mathrm{d}t=\frac{1}{2i}\int_0^5e^{(-2+i)t}-e^{(-2-i)t}\mathrm{d}t=\frac{1}{2i}\left.\left(\frac{e^{(-2+i)t}}{-2+i}-\frac{e^{(-2-i)t}}{2-i}\right)\right|_0^5= $$ $$ =\frac{(-2-i)e^{(-2+i)t}-(-2+i)e^{(-2-i)t}}{10i}=\frac{1-\mathrm e^{-10}(\cos 5+2\sin 5)}5. $$

asomog
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