If int x^3 sin 3x , dx = f(x) cos 3x + g(x) s | vedclass

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Question

(C) Using integration by parts $\int u \, dv = uv - \int v \, du$ repeatedly:$\int x^3 \sin 3x \, dx = x^3 \left( \frac{-\cos 3x}{3} \right) - \int 3x

Vedclass · Accepted Answer

$4x$

Vedclass · Answer

(C) Using integration by parts $\int u \, dv = uv - \int v \, du$ repeatedly:$\int x^3 \sin 3x \, dx = x^3 \left( \frac{-\cos 3x}{3} \right) - \int 3x^2 \left( \frac{-\cos 3x}{3} \right) dx = -\frac{x^3 \cos 3x}{3} + \int x^2 \cos 3x \, dx$$= -\frac{x^3 \cos 3x}{3} + \left( x^2 \frac{\sin 3x}{3} - \int 2x \frac{\sin 3x}{3} \, dx \right)$$= -\frac{x^3 \cos 3x}{3} + \frac{x^2 \sin 3x}{3} - \frac{2}{3} \int x \sin 3x \, dx$Now, $\int x \sin 3x \, dx = x \left( \frac{-\cos 3x}{3} \right) - \int 1 \left( \frac{-\cos 3x}{3} \right) dx = -\frac{x \cos 3x}{3} + \frac{\sin 3x}{9}$Substituting back:$\int x^3 \sin 3x \, dx = -\frac{x^3 \cos 3x}{3} + \frac{x^2 \sin 3x}{3} - \frac{2}{3} \left( -\frac{x \cos 3x}{3} + \frac{\sin 3x}{9} \right)$$= -\frac{x^3 \cos 3x}{3} + \frac{x^2 \sin 3x}{3} + \frac{2x \cos 3x}{9} - \frac{2 \sin 3x}{27}$$= \cos 3x \left( -\frac{x^3}{3} + \frac{2x}{9} \right) + \sin 3x \left( \frac{x^2}{3} - \frac{2}{27} \right) + c$Comparing with $f(x) \cos 3x + g(x) \sin 3x + c$, we get $f(x) = -\frac{x^3}{3} + \frac{2x}{9}$ and $g(x) = \frac{x^2}{3} - \frac{2}{27}$Then $27(f(x) + x g(x)) = 27 \left( -\frac{x^3}{3} + \frac{2x}{9} + x \left( \frac{x^2}{3} - \frac{2}{27} \right) \right)$$= 27 \left( -\frac{x^3}{3} + \frac{2x}{9} + \frac{x^3}{3} - \frac{2x}{27} \right) = 27 \left( \frac{6x - 2x}{27} \right) = 4x$

If $\int x^3 \sin 3x \, dx = f(x) \cos 3x + g(x) \sin 3x + c$, then $27(f(x) + x g(x)) =$

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