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I tried to solve the following question by integration by parts but it gets iterative and no solution is found. $$\int e^{-x}(\cos wx + w\sin wx)dx$$ where $w$ is constant.

Is there any method by which this question could be solved? I don't want the whole solution but an idea how to deal with such type of questions.

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    $\begingroup$ Hint: Compute the derivative of the functions $f(x)=e^{-x}\cos(wx)$ and $g(x)=e^{-x}\sin(wx)$ in terms of $f$ and $g$ and deduce expressions of $f$ and $g$ as derivatives. $\endgroup$ – Did Oct 6 '15 at 10:37
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Hint; Separate the intgral into 2 parts like this;

$$\int e^{-x}(\cos wx + w\sin wx)dx= \int e^{-x}(\cos wx) dx+\int e^{-x}(w\sin wx)dx$$

Then integrate by parts the first one 'till magic happens.

Hint#2; $e^{-x}\cos wx$ can be written as $-\cos wx \times -e^{-x}$

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Forget integration by parts here. There's a simpler way. I'll do one of the terms, the other is similar.

$$\int e^{-x}\cos{wx} dx = Re(\int e^{-x}\cos{wx} dx) = Re(\int e^{-(1+jw)x} dx) = Re(-1/(1+jw) e^{-(1+jw)x}) = -1/(1-w^2)e^{-x}\cos{wx}$$

Double check my algebra please - I haven't done this in years.

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  • $\begingroup$ what is Re? (1+jw) came from where? $\endgroup$ – Sohail Ahmed Oct 6 '15 at 11:30
  • $\begingroup$ Re() is a function which returns the real part of a complex number. Re(a+jb) = a. Some people use i for the imaginary unit. 1/(1+jw) came from integrating the exponent. $\endgroup$ – s5s Oct 6 '15 at 12:12
  • $\begingroup$ I didn't understand the concept where did coswx go.You multiply the exponent of e with 1+jw......? $\endgroup$ – Sohail Ahmed Oct 6 '15 at 12:50
  • $\begingroup$ This is what happened: $$re^{jwx} = r(\cos(wx) + j\sin(wx))$$ where r, w are real numbers.Then of course, $$Re(re^{jwx}) = r\cos(wx)$$ $\endgroup$ – s5s Oct 6 '15 at 13:10
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One can note that

$$\int e^{-x}(\cos wx + w\sin wx)dx = \int e^{g(x)}[f'(x) + g'(x)f(x)] dx$$

with $f(x)= - \cos wx $ and $g(x)=-x.$

So according the the formula $$\int e^{g(x)}[f'(x) + g'(x)f(x)] dx = f(x) e^{g(x)}+C$$ the answer is
$$\int e^{-x}(\cos wx + w\sin wx)dx =f(x)e^{g(x)} = -(\cos wx) e^{-x}+C.$$

For the proof of the formula one can see:

Proof for formula $\int e^{g(x)}[f'(x) + g'(x)f(x)] dx = f(x) e^{g(x)}+C$

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