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I know that using the product rule, I can calculate the derivative of the following function:

$$[f(x)\cos(\omega x)]' = f'(x)\cos(\omega x)-\omega f(x)\sin(\omega x) $$

However, how could I find the anti-derivative of the following function, if $k$ is a real number (possibly $-1\leq k \leq 1 $):

$$ kf'(x)\cos(\omega x)-\omega f(x)\sin(\omega x) $$ ??

Thanks in advance!

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    Did you mean to say product rule? The reverse of product rule is integration by parts, it that's what you are looking for. Without more information, you can't do this for general $k$. – 2017-02-18
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    Yep! I just corrected that ;) Which restrictions should apply for k, in order to solve that... Isn't it sufficient to limit k as less than 1? – 2017-02-18
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    I meant that you can't find the anti-derivative for $k\ne1$ without knowing what $f$ is. – 2017-02-18
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    Ah, ok, I get it. The function f is an exponential in the form $$ K_0 +(K_1-K_0)e^{(-\alpha x)} $$ – 2017-02-18
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    Well, that could be the case. Indeed, that is the simplest case, but if $k=1$, we may find the antiderivative even without $f$ being an exponential function and all that. – 2017-02-18

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