Integrating using by part, we get, #160; v = e^ax , u = sin bx ∫ v = 1a e^ax , u^' = b cos bx #160; e^axsin bxa ∫b e^axcos bx dxa #160;A ...

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Integration by Parts

Solve ∫ eax...

Question

Solve $\int e^{a x} sin b x d x$ .

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Solution

Integrating using by part, we get,
$v = e^{a x}, u = sin b x$
$\int v = \frac{1}{a} e^{a x}, u^{^{'}} = b cos b x$
$\frac{e^{a x} sin b x}{a} - \int \frac{b e^{a x} cos b x d x}{a}$
Again using by parts we get,
$\frac{e^{a x} sin b x}{a} - \frac{b e^{a x} cos b x}{a^{2}} - \int \frac{b^{2} e^{a x} sin b x d x}{a^{2}}$
$\frac{(a^{2} + b^{2}) I}{a^{2}} = \frac{e^{a x} sin b x}{a} - \frac{b e^{a x} cos b x}{a^{2}}$
$I = \frac{1}{a^{2} + b^{2}} (a e^{a x} sin b x - b e^{a x} cos b x)$

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