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How to integrate $$\int_{0}^{\pi/2} e^{−2x}\sin(3x)\rm dx $$

I have attempted to this question with integration by parts, but I'm hitting a lot of walls. I have a feeling it might have to do with representing repeating integrals with I, but I'm not quite sure as to how to approach this question. Sorry for the terrible formatting, it's my first time on this site.

user121167
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4 Answers4

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Letting $I$ be your definite integral,

$$\begin{align}I=\int_{0}^{\pi/2}e^{-2x}\sin(3x)dx&=\int_{0}^{\pi/2}\left(-\frac 12e^{-2x}\right)^\prime\sin (3x)dx\\&=(-1/2)e^{-2x} \sin(3x) - \int_{0}^{\pi/2}(-1/2)e^{-2x} \cdot 3\cdot \cos(3x) dx\\&=(-1/2)e^{-2x} \sin(3x) -(-3/2)\int_{0}^{\pi/2} e^{-2x} \cos(3x) dx.\end{align}$$

Also, since $$\begin{align}\int_{0}^{\pi/2} e^{-2x} \cos(3x) dx　&= (-1/2)e^{-2x} \cos(3x) - \int_{0}^{\pi/2} (-1/2)e^{-2x} (-3) \sin(3x) dx\\&=(-1/2)e^{-2x} \cos(3x) -(3/2)I,\end{align}$$ you'll get $$I= (-1/2)e^{-2x} \sin(3x) - (-3/2)\{ (-1/2)e^{-2x} \cos(3x) - (3/2)I \}.$$ Then, you can solve this equation of $I$.

mathlove
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Hint: Use the fact that $\sin x=\Im(e^{ix})$, or $\sin x=\dfrac{e^{ix}-e^{-ix}}2$ .

Lucian
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  • This gets you an antiderivative that is not immediately easy to convert back in to an $i$-free form, if that is a concern. – 2'5 9'2 Jan 14 '14 at 06:09
  • @alex.jordan: What are you talking about ? You get $I=\Im\dfrac{e^{(a+ib)x}}{a+ib}$ . Just rewrite the numerator as $e^{ax}(\cos bx+i\sin bx)$, and then amplify with the conjugate of the denominator, $a-bi$, select the imaginary part, and you're done! – Lucian Jan 14 '14 at 06:25
  • Right; it's not immediate. Selecting the imaginary part still has some steps to it. OP's question strikes me as a standard question from an introductory integral calculus class. I'll let OP weight in, but I would doubt (s)he knows about imaginary parts of $e^{ix}$ and the like. – 2'5 9'2 Jan 14 '14 at 07:16
  • Haha thanks for the comment Lucian. Unfortunately the scope of my course doesn't cover into imaginary numbers and such, but I'm sure your method works just as well as the integration by parts method. – user121167 Jan 14 '14 at 07:35
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First you want to find a function $y(x)$ such that $y'(x)=e^{-2x}\sin3x$. For this you can use the method of undetermined coefficients. To start with, you guess (or know) that there is a solution of the form $y=Ae^{-2x}\cos3x+Be^{-2x}\sin3x$ where $A$ and $B$ are constants to be determined. Differentiate and equate coefficients: $$y=e^{-2x}(A\cos3x+B\sin3x);$$ $$y'=-2e^{-2x}(A\cos3x+B\sin3x)+e^{-2x}(-3A\sin3x+3B\cos3x)=e^{-2x}(-2A+3B)\cos3x+e^{-2x}(-3A-2B)\sin3x=e^{-2x}\sin3x;$$ $$-2A+3B=0,\ -3A-2B=1.$$ The solution of the linear equations is $A=\frac{-3}{13}$ and $B=\frac{-2}{13}$ (if I did all that arithmetic right, which is not very likely), so your antiderivative is $$y(x)=\frac1{13}e^{-2x}(-3\cos3x-2\sin3x)$$ and the value of your definite integral is $$\int_0^\frac\pi2e^{-2x}\sin3xdx=y(\frac\pi2)-y(0)=\frac{2e^{-\pi}+3}{13}$$ or something like that. Alternatively, you could just look up the formula for $\int e^{ax}\sin bxdx$.

bof
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You have to be careful using integration by parts:

$$\int e^{-2x} \sin (3x) dx= -\frac{1}{2} e^{-2x} \sin (3x) +\frac{3}{2} \int e^{-2x} \cos (3x) dx = -\frac{1}{2} e^{-2x} \sin (3x) -\frac{3}{4} e^{-2x} \cos (3x)-\frac{9}{4} \int e^{-2x} \sin (3x) dx$$ note that in the latter part you recovered the initial integral (but with a different coefficient). Thus

$$\int e^{-2x} \sin (3x) dx= -\frac{2}{13} e^{-2x} \sin (3x)-\frac{3}{13} e^{-2x} \cos (3x)$$ and

$$\int_{0}^{\frac{\pi}{2}} e^{-2x} \sin (3x) dx= -\frac{2}{13}(e^{-\pi}\sin(\frac{3\pi}{2})-e^{0} \sin(0)) -\frac{3}{13}(e^{-\pi}\cos(\frac{3\pi}{2})-e^{0} cos(0))=\frac{2}{13}e^{-\pi}+\frac{3}{13}$$

Matheman
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