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Higher Order Derivatives

If ∫ e 2x s...

Question

If $\int e^{2 x} sin 3 x d x = \frac{e^{2 x}}{A} (2 sin 3 x - 3 cos 3 x) + c$ , then $A$ is

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Solution

We have, $\int u . v d x = u \int v d x$ $- \int (\frac{d u}{d x} \int v d x) d x$ .......... Integration by parts

Let $I = \int e^{2 x} sin 3 x d x$

$= e^{2 x} (- \frac{cos 3 x}{3}) - \int 2 e^{2 x} (- \frac{cos 3 x}{3}) d x$ .......... Integration by parts

$= - \frac{1}{3} e^{2 x} cos 3 x + \frac{2}{3} \int e^{2 x} cos 3 x d x$

$= - \frac{1}{3} e^{2 x} cos 3 x + \frac{2}{3} [e^{2 x} \frac{sin 3 x}{3} - \int 2 e^{2 x} \frac{sin 3 x}{3} d x]$ .......... Integration by parts

$= - \frac{1}{3} e^{2 x} cos 3 x + \frac{2}{9} e^{2 x} sin 3 x - \frac{4}{9} \int e^{2 x} sin 3 x d x$

$= - \frac{1}{3} e^{2 x} cos 3 x + \frac{2}{9} e^{2 x} sin 3 x - \frac{4}{9} I$

$\Rightarrow I + \frac{4}{9} I = \frac{e^{2 x}}{9} (2 sin 3 x - 3 cos 3 x) \Rightarrow \frac{13}{9} I = \frac{e^{2 x}}{9} (2 sin 3 x - 3 cos 3 x)$

$\Rightarrow I = \frac{e^{2 x}}{13} (2 sin 3 x - 3 cos 3 x) + c$

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