Use a table of integrals with forms involving eu to find the indefinite integral ∫e−4xsin⁡3xdx

Question

braodagxj · Accepted Answer

Step 1
By using the table of integrals we are to find the indefinite integral:

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

Step 2
We will use formula from the table of integrals with forms involving

e^{u}

:

\int e^{u x} \sin v x d x = \frac{e^{u x}}{u^{2} + v^{2}} [u \sin v x - v \cos v x] + C

...(1)
We are to find the indefinite integral

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

So, Here u=-4 and v=3
On Putting u=-4 and v=3 in (1) we get:

\int e^{- 4 x} \sin 3 x d x = \frac{e^{- 4 x}}{{(- 4)}^{2} + {(3)}^{2}} [- 4 \sin 3 x - 3 \cos 3 x] + C

\int e^{- 4 x} \sin 3 x d x = \frac{e^{- 4 x}}{25} [- 4 \sin 3 x - 3 \cos 3 x] + C

or

\int e^{- 4 x} \sin 3 x d x = - \frac{e^{- 4 x}}{25} [4 \sin 3 x + 3 \cos 3 x] + C

lenkiklisg7 · Accepted Answer

∫e−4xsin⁡(3x)dx  The formula for integration by parts:  ∫U(x)⋅dV(x)=U(x)⋅V(x)−∫V(x)⋅dU(x)  We put  U=1  dV=dx  Then:  dU=0dx  V=x  Therefore:  ∫e−4x⋅sin⁡(3x)dx=x  Answer:  ∫e−4x⋅sin⁡(3x)=x+C

Vasquez · Accepted Answer

∫e−4xsin⁡(3x)dx=−e−4xsin⁡(3x)4−∫−3e−4xcos⁡(3x)4dx=−e−4xsin⁡(3x)4−(3e−4xcos⁡(3x)16−∫−9e−4xsin⁡(3x)16dx)=−e−4xsin⁡(3x)4−(3e−4xcos⁡(3x)16+916∫e−4xsin⁡(3x)dx)=−4e−4xsin⁡(3x)−3e−4xcos⁡(3x)25∫e−4xsin⁡(3x)dx=−4e−4xsin⁡(3x)−3e−4xcos⁡(3x)25+C=−e−4x(4sin⁡(3x)+3cos⁡(3x))25+C

Use a table of integrals with forms involving eu to

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