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I have this Integration by Parts question that I can't seem to find an answer to.

The question is:

$$\int\ e^x \sin(9x)\,dx$$

I used u-substitution:

$$u=e^x,du=e^x\,dx$$ $$dv=\sin(9x)\,dx, v=-\frac{1}{9}\cos(9x)$$

Then I got:

$$-\frac{1}{9}e^x\cos(9x)+\frac{1}{9}\int\ e^x\cos(9x)\,dx$$

After a second integration I used:

$$u=e^x, du=e^x\,dx$$

$$dv=\cos(9x)\,dx, v=\frac{1}{9}\sin(9x)$$

Furthermore,

$$\int\ e^x\sin(9x)dx=-\frac{1}{9}\ e^x\cos(9x)+\frac{1}{9}\left(\frac{1}{9}\ e^x\sin(9x)-\frac{1}{9}\int\ e^x\sin(9x)\,dx\right)$$

$$\int\ e^x\sin(9x)dx=-\frac{1}{9}\ e^x\cos(9x)+\frac{1}{81}\ e^x\sin(9x)-\frac{1}{81}\int\ e^x\sin(9x)\,dx$$

I'm stuck and I'm not sure exactly what to do after this.

Any help would be very grateful..thanks!

Martin Sleziak
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AlecLeonK
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    For clear expression of any trigonometric equations, be sure to include \ before cosine and sine. – NasuSama Mar 08 '14 at 01:04

4 Answers4

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More simply and without integration by parts we have $$\int e^x\sin(9x)dx=\operatorname{Im}\int e^{(1+9i)x}dx=\operatorname{Im}\left(\frac1{1+9i}e^{(1+9i)x}\right)=\operatorname{Im}\left(\frac1{82}(1-9i)e^{(1+9i)x}\right)$$

Now develop and take the imaginary part.

NasuSama
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  • Where did you learn this? I've never seen it answered like that before – AlecLeonK Mar 08 '14 at 01:22
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    @AlecLeonK First, write $e^{(1 + 9i)x}$ as $e^xe^{9ix}$. Then, for $e^{9ix}$, use Euler's identity and find the imaginary part of that expression. For more information about this method, see here. – NasuSama Mar 08 '14 at 01:31
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From where you left off, let

$$s = \int e^x\sin(9x)\,dx$$

Then, we have

$$s = -\dfrac{1}{9}e^x\cos(9x) + \dfrac{1}{81}e^x\sin(9x) - \dfrac{1}{81}s$$

Solving for $s$ gives

$$\begin{aligned} \dfrac{82}{81}s &= -\dfrac{1}{9}e^x\cos(9x) + \dfrac{1}{81}e^x\sin(9x)\\ s &= \dfrac{81}{82}\left(-\dfrac{1}{9}e^x\cos(9x) + \dfrac{1}{81}e^x\sin(9x) \right)\\ \int e^x\sin(9x)\,dx &= \dfrac{81}{82}\left(-\dfrac{1}{9}e^x\cos(9x) + \dfrac{1}{81}e^x\sin(9x) \right) \end{aligned}$$

NasuSama
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  • I typed it in and I see the logic behind it but for some reason it says it's wrong. I understand what you did though. – AlecLeonK Mar 08 '14 at 01:11
  • I'm very glad to hear that. I think you ask this question that comes from the online homework, am I right? – NasuSama Mar 08 '14 at 01:22
  • Yes it does, do you know why it might be wrong? – AlecLeonK Mar 08 '14 at 01:40
  • I don't know. Assuming your online HW system does not give away answer after exhausted amount of attempts or after HW deadline, I would assume that the HW system computed the integral in the complicated way without using the elementary method you learned (I guessed). – NasuSama Mar 08 '14 at 01:42
  • Would it make a difference if I make the orginal 'u'=sin(9x) instead of e^x? – AlecLeonK Mar 08 '14 at 01:46
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We can take a more general approach, and find $$I=\int e^{ax}\sin bx\ \mathrm{d}x$$ Integration by parts: $$\mathrm{d}v=\sin bx\ \mathrm{d}x\Rightarrow v=-\frac1b\cos bx\\u=e^{ax}\Rightarrow \mathrm{d}u=ae^{ax}\mathrm{d}x$$ Plug it in: $$I=-\frac{e^{ax}}b\cos bx+\frac ab\int e^{ax}\cos bx\ \mathrm{d}x$$ Integration by parts round $2$: $$\mathrm{d}v=\cos bx\ \mathrm{d}x\Rightarrow v=\frac1b\sin bx\\u=e^{ax}\Rightarrow \mathrm{d}u=ae^{ax}\mathrm{d}x$$ Plug in: $$I=-\frac{e^{ax}}b\cos bx+\frac ab\bigg(\frac{e^{ax}}{b}\sin bx-\frac ab\int e^{ax}\sin bx\ \mathrm{d}x\bigg)$$ Take note of the $\int e^{ax}\sin bx\ \mathrm{d}x$ term: $$I=-\frac{e^{ax}}b\cos bx+\frac{ae^{ax}}{b^2}\sin bx-\frac{a^2}{b^2}I$$ $$\bigg(1+\frac{a^2}{b^2}\bigg)I=\frac{e^{ax}}b\bigg(\frac ab\sin bx-\cos bx\bigg)$$ $$I=\frac{b^2}{a^2+b^2}\frac{e^{ax}}b\bigg(\frac ab\sin bx-\cos bx\bigg)$$ $$I=\frac{e^{ax}}{a^2+b^2}\big(a\sin bx-b\cos bx\big)+C$$ Isn't that cool? This is called a reduction formula, and there are many more of them.

clathratus
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You have $$\int e^x\sin(9x)dx=−\frac{1}{9}e^x\cos(9x)+\frac{1}{81} e^x\sin(9x)−\frac{1}{81}\int e^x\sin(9x)dx$$ Then bring $−\frac{1}{81}\int e^x\sin(9x)dx$ to the other side and you get $$\frac{82}{81}\int e^x\sin(9x)dx=−\frac{1}{9} e^x\cos(9x)+\frac{1}{81} e^x\sin(9x)$$

AlecLeonK
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Kaladin
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