int e^{-3 x}left(x^2+sin 4 xright) d x= | vedclass

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(A) We need to evaluate $I = \int e^{-3x}(x^2 + \sin 4x) dx = \int x^2 e^{-3x} dx + \int e^{-3x} \sin 4x dx$. For the first part, using integration by

Vedclass · Accepted Answer

$-e^{-3 x}\left(\frac{x^2}{3}+\frac{2 x}{9}+\frac{2}{27}+\frac{3}{25} \sin 4 x+\frac{4}{25} \cos 4 x\right)+C$

Vedclass · Answer

(A) We need to evaluate $I = \int e^{-3x}(x^2 + \sin 4x) dx = \int x^2 e^{-3x} dx + \int e^{-3x} \sin 4x dx$. For the first part, using integration by parts $\int u dv = uv - \int v du$ with $u = x^2$ and $dv = e^{-3x} dx$: $\int x^2 e^{-3x} dx = x^2 \left(\frac{e^{-3x}}{-3}\right) - \int 2x \left(\frac{e^{-3x}}{-3}\right) dx = -\frac{x^2}{3} e^{-3x} + \frac{2}{3} \int x e^{-3x} dx$. Applying integration by parts again: $\int x e^{-3x} dx = x \left(\frac{e^{-3x}}{-3}\right) - \int 1 \left(\frac{e^{-3x}}{-3}\right) dx = -\frac{x}{3} e^{-3x} - \frac{1}{9} e^{-3x}$. So, $\int x^2 e^{-3x} dx = -\frac{x^2}{3} e^{-3x} - \frac{2}{9} x e^{-3x} - \frac{2}{27} e^{-3x}$. For the second part, using the formula $\int e^{ax} \sin bx dx = \frac{e^{ax}}{a^2+b^2} (a \sin bx - b \cos bx)$: $\int e^{-3x} \sin 4x dx = \frac{e^{-3x}}{(-3)^2 + 4^2} (-3 \sin 4x - 4 \cos 4x) = -\frac{e^{-3x}}{25} (3 \sin 4x + 4 \cos 4x)$. Combining both parts: $I = -e^{-3x} \left( \frac{x^2}{3} + \frac{2x}{9} + \frac{2}{27} + \frac{3}{25} \sin 4x + \frac{4}{25} \cos 4x \right) + C$.

$\int e^{-3 x}\left(x^2+\sin 4 x\right) d x=$

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