If int x^3 sin 3x , dx = frac{1}{27}[f(x) cos | vedclass

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(C) We use integration by parts: $\int u \, dv = uv - \int v \, du$.Let $I = \int x^3 \sin 3x \, dx$.Using the tabular method for integration by parts

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$4$

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(C) We use integration by parts: $\int u \, dv = uv - \int v \, du$.Let $I = \int x^3 \sin 3x \, dx$.Using the tabular method for integration by parts:$u = x^3$, $dv = \sin 3x \, dx$$u' = 3x^2$, $v = -\frac{1}{3} \cos 3x$$u'' = 6x$, $v_1 = -\frac{1}{9} \sin 3x$$u''' = 6$, $v_2 = \frac{1}{27} \cos 3x$$u'''' = 0$, $v_3 = \frac{1}{81} \sin 3x$$I = (x^3)(-\frac{1}{3} \cos 3x) - (3x^2)(-\frac{1}{9} \sin 3x) + (6x)(\frac{1}{27} \cos 3x) - (6)(\frac{1}{81} \sin 3x) + c$$I = -\frac{x^3}{3} \cos 3x + \frac{x^2}{3} \sin 3x + \frac{2x}{9} \cos 3x - \frac{2}{27} \sin 3x + c$Multiply by $\frac{27}{27}$ to match the form $\frac{1}{27}[f(x) \cos 3x + g(x) \sin 3x]$:$I = \frac{1}{27}[-9x^3 \cos 3x + 9x^2 \sin 3x + 6x \cos 3x - 2 \sin 3x] + c$$I = \frac{1}{27}[(-9x^3 + 6x) \cos 3x + (9x^2 - 2) \sin 3x] + c$Thus, $f(x) = -9x^3 + 6x$ and $g(x) = 9x^2 - 2$.$f(1) = -9(1)^3 + 6(1) = -9 + 6 = -3$.$g(1) = 9(1)^2 - 2 = 9 - 2 = 7$.$f(1) + g(1) = -3 + 7 = 4$.

If $\int x^3 \sin 3x \, dx = \frac{1}{27}[f(x) \cos 3x + g(x) \sin 3x] + c$, then $f(1) + g(1) =$

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