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Solve $\int x^2e^x\sin x$

My attempt is as follows:-

$$I_1=\int e^x\sin x$$ $$I_1=e^x\sin x-\int e^x\cos x$$ $$I_1=e^x\sin x-e^x\cos x-\int\sin (x)e^x$$ $$2I_1=e^x\left(\sin x-\cos x\right)$$ $$I_1=\dfrac{e^x\left(\sin x-\cos x\right)}{2}\tag{1}$$

$$I_2=\int e^x\cos x$$ $$I_2=e^x\cos x+\int e^x\sin x$$ $$I_2=e^x\cos x+e^x\sin x-\int e^x \cos x$$ $$I_2=\dfrac{e^x(\cos x+\sin x)}{2}$$

$$I=\int x^2e^x\sin x$$ $$I=I_1x^2-2\int xI_1$$ $$I=I_1x^2-\int xe^x(\sin x-\cos x)$$ $$I=I_1x^2-\int xe^x\sin x+\int xe^x\cos x$$

$$I=I_1x^2-xI_1+\int I_1+xI_2-\int I_2$$ $$I=I_1x^2-xI_1+xI_2+\int I_1-\int I_2$$ $$I=I_1x^2-xI_1+xI_2+\dfrac{1}{2}\int e^x\left(\sin x-\cos x\right) -\dfrac{1}{2}\int e^x\left(\cos x+\sin x\right)$$

$$I=I_1x^2-xI_1+xI_2+\dfrac{I_1}{2}-\dfrac{I_2}{2} -\dfrac{I_2}{2}-\dfrac{I_1}{2}$$

$$I=I_1x^2-xI_1+xI_2-I_2$$ $$\dfrac{e^x}{2}\left(x^2\sin x-x^2\cos x-x\sin x+x\cos x+x\cos x+x\sin x-\sin x-\cos x \right)+C$$

$$\dfrac{e^x}{2}\left((x^2-1)\sin x-(x-1)^2\cos x \right)+C$$

Is there any better way to solve it which is short and clean. Mine got very long.

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  • $\begingroup$ How about $\sin x=\Im e^{i x}$ $\endgroup$ – Ali Shather Dec 24 '19 at 19:46
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    $\begingroup$ no we can't use complex numbers. $\endgroup$ – user3290550 Dec 24 '19 at 19:47
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    $\begingroup$ I think you're forgetting a few integration constants here and there. $\endgroup$ – Arthur Dec 24 '19 at 19:49
  • $\begingroup$ You can get rid of the $i$ at the end. $\endgroup$ – Ali Shather Dec 24 '19 at 19:50
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    $\begingroup$ An obvious way to save a line is to note $$I_2=e^x \sin(x)-I_1$$ so you don't need to recalculate after having calculated $I_1$ $\endgroup$ – Maximilian Janisch Dec 24 '19 at 19:52
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I don't know if you like this or not. Let $$ I(a)=\int e^{ax}\sin(x)dx. $$ It is easy to see $$ I''(a)=\int x^2 e^{ax}\sin(x)dx. $$ But $$ I(a)=\frac{e^{ax}(-\cos x+a\sin x)}{a^2+1}+C(a). $$ Now taking the 2nd derivative will give the answer.

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Start with writing $\sin x=\color{red}{\Im}e^{ix}$

$$\int x^2e^x \sin x\ dx=\color{red}{\Im}\int x^2e^{(1+i)x}\ dx$$

$$=\color{red}{\Im}\left(\frac{x^2}{1+i}e^{(1+i)x}-\frac{2x}{(1+i)^2}e^{(1+i)x}+\frac{2}{(1+i)^3}e^{(1+i)x}\right)$$

$$=x^2 \color{red}{\Im} \frac{e^{(1+i)x}}{1+i}-2x \color{red}{\Im}\frac{e^{(1+i)x}}{(1+i)^2}+2\color{red}{\color{red}{\Im}} \frac{e^{(1+i)x}}{(1+i)^3}$$

$$=x^2\left(\frac12e^x\sin x-\frac12e^x\cos x\right)-2x\left(-\frac12e^x\cos x\right)+2\left(-\frac14e^x\sin x-\frac14e^x\cos x\right)$$

$$=\frac12e^x\sin x(x^2-1)-\frac12e^x\cos x(x-1)^2+C$$

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  • $\begingroup$ by the way your answer is wrong. $\endgroup$ – user3290550 Dec 24 '19 at 20:05
  • $\begingroup$ yes i know what i did wrong. thanks $\endgroup$ – Ali Shather Dec 24 '19 at 20:09

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