Using integration by parts nbsp; I=∫ e^ 3xsin4xdx = e^ 3xcos 4x4 ∫34 e^ 3xcos 4x dx = e^ 3xcos 4x4 34(e^ 3xsin 4x4+∫ 3e^ 3xsin 4x4dx) ⇒25I16= e^ 3xcos 4x4 ...

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Integration by Parts

∫ e-3xsin4xdx...

Question

$\int e^{- 3 x} sin 4 x d x =$

\frac{e^{- 3 x} (4 cos 4 x - 3 sin 4 x)}{25} + c

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\frac{- e^{- 3 x} (3 cos 4 x + 4 sin 4 x)}{25} + c

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\frac{- e^{- 3 x} (4 cos 4 x + 3 sin 4 x)}{25} + c

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None of these

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Solution

The correct option is C $\frac{- e^{- 3 x} (4 cos 4 x + 3 sin 4 x)}{25} + c$
Using integration by parts
$I = \int e^{- 3 x} sin 4 x d x = - e^{- 3 x} \frac{cos 4 x}{4} - \int \frac{3}{4} e^{- 3 x} cos 4 x d x = - e^{- 3 x} \frac{cos 4 x}{4} - \frac{3}{4} (e^{- 3 x} \frac{sin 4 x}{4} + \int 3 e^{- 3 x} \frac{sin 4 x}{4} d x) \Rightarrow \frac{25 I}{16} = - e^{- 3 x} \frac{cos 4 x}{4} - \frac{3}{16} e^{- 3 x} sin 4 x + c \Rightarrow I = \frac{- e^{- 3 x} (4 cos 4 x + 3 sin 4 x)}{25} + c$

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∫e−3xsin4xdx=

The correct option is C −e−3x(4cos4x+3sin4x)25+cUsing integration by parts I=∫e−3xsin4xdx=−e−3xcos4x4−∫34e−3xcos4xdx=−e−3xcos4x4−34(e−3xsin4x4+∫3e−3xsin4x4dx)⇒25I16=−e−3xcos4x4−316e−3xsin4x+c⇒I=−e−3x(4cos4x+3sin4x)25+c

$\int e^{- 3 x} sin 4 x d x =$