Mathematics

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


SOLUTION
$$I=\displaystyle\int x^{2}e^{x}\sin xdx$$
Integrating by parts $$u=x^{2}. V=e^{x}\sin x$$
$$u^{1}=2x, \displaystyle\int vdx=\dfrac{1}{2}e^{x} (\sin x-\cos x)$$  [Integrated by parts]
$$I=x^{2}\displaystyle\int e^{x}\sin dx-\int 2x \left(\int e^{x}\sin x dx\right)dx$$
$$=\dfrac{x^{2}e^{x}}{2}(\sin x-\cos x)-\int xe^{x}(\sin x-\cos x)dx$$
Now, $$\displaystyle\int e^{x}(\cos x-\sin x)dx=e^{x}\cos x [\because \int e^{x}f(x)+f'(x)dx=e^{x}f(x)]$$
$$I=\dfrac{x^{2}e^{x}}{2}(\sin x-\cos x)+\left[x e^{x}\cos x-\int (1) e^{x}\cos xdx\right]$$     [Using by parts]
$$=\dfrac{x^{2}e^{x}}{2}(\sin x-\cos x)+xe^{x}\cos x-\dfrac{e^{x}}{2} (\sin x+\cos x)+C$$ [Using by parts]
$$=\dfrac{e^{x}}{2}[\sin (x)(x^{2}-1)+\cos x(-x^{2}+2x-1)]+C$$
$$I=\dfrac{e^{x}}{2}((x^{2}-1)\sin x-(x-1)^{2}\cos x)+C$$
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Subjective Medium Published on 17th 09, 2020
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