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What are $$\int x e^{ax} \cos(bx)dx$$ and $$\int x e^{ax} \sin(bx)dx?$$

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    These are senseless comments. Please expand on what you are trying to say. – 2017-01-18
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    Use integration by parts. Remember the product rule? – 2017-01-18
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    @spaceisdarkgreen I think he just means the indefinite integral evaluated for $a\ne 0$ and $b\ne 0$ (the non-trivial solutions) – 2017-01-18

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One can see that

$$\int e^{ax}\cos(bx)\ dx=\Re\int e^{(a+bi)x}\ dx=\frac{e^{ax}}{a^2+b^2}\left(a\cos(bx)+b\sin(bx)\right)+c$$

Likewise,

$$\int e^{ax}\sin(bx)\ dx=\Im\int e^{(a+bi)x}\ dx=\frac{e^{ax}}{a^2+b^2}\left(a\sin(bx)-b\cos(bx)\right)+c$$

And lastly differentiate each w.r.t. $a$ to reveal

$$\int xe^{ax}\cos(bx)\ dx=\frac{e^{ax}}{a^2+b^2}\left((ax+1-\frac{2a}{a^2+b^2})\cos(bx)+(bx-\frac{2a}{a^2+b^2})\sin(bx)\right)+c$$

$$\int e^{ax}\sin(bx)\ dx=\frac{e^{ax}}{a^2+b^2}\left((ax+1-\frac{2a}{a^2+b^2})\sin(bx)-(bx+\frac{2a}{a^2+b^2})\cos(bx\right)+c$$

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    Great! Thank you very much! – 2017-01-18
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    @regress Sure, your welcome. :D – 2017-01-18
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    (honestly did it this way just to avoid integration by parts, since that is probably the normal method) – 2017-01-18
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Integration by parts for 3 functions -

$$\int fgh'\ dx=fgh-\int fg'h+f'gh\ dx$$

Taking reference to this link you can solve above integrals.

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    Your welcome... – 2017-01-18
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    T_T this problem was made to kill my soul. Triple integration by parts? Noooppppeeeeee! (You could do the integration by parts on just $e^{ax}\cos(bx)$ like I did and then just differentiate like I did.) – 2017-01-18
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    Hahaha... Its very difficult to solve triple integrals by parts. And sometimes irritated me also. And link I provided is very nice. – 2017-01-18
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    Yeah. Personally, I've never been forced into IBP due to methods like above. – 2017-01-18