How do you find the antiderivative of #e^(2x) * sin(4x)dx#?
OR
Here,
We know that,
#color(red)((1)inte^(ax) sinbxdx=e^(ax)/(a^2+b^2)(asinbx- bcosbx)+c#
So,
OR
We also know that,
#color(red)((2)inte^(ax)sinbxdx=e^(ax)/sqrt(a^2+b^2)sin(bx- theta)+c#,
So,
Hence,
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To find the antiderivative of ( e^{2x} \cdot \sin(4x) ) with respect to ( x ), you can use integration by parts. Let ( u = e^{2x} ) and ( dv = \sin(4x)dx ). Then, ( du = 2e^{2x}dx ) and ( v = -\frac{1}{4}\cos(4x) ). Applying the integration by parts formula:
[ \int e^{2x} \cdot \sin(4x)dx = -\frac{1}{4}e^{2x}\cos(4x) - \frac{1}{2} \int e^{2x} \cdot \cos(4x)dx ]
To evaluate the remaining integral, you can use integration by parts again. Let ( u = e^{2x} ) and ( dv = \cos(4x)dx ). Then, ( du = 2e^{2x}dx ) and ( v = \frac{1}{4}\sin(4x) ). Applying the integration by parts formula:
[ \int e^{2x} \cdot \cos(4x)dx = \frac{1}{4}e^{2x}\sin(4x) - \frac{1}{2} \int e^{2x} \cdot \sin(4x)dx ]
Rearranging terms, we get:
[ \frac{3}{2}\int e^{2x} \cdot \sin(4x)dx = -\frac{1}{4}e^{2x}\cos(4x) + \frac{1}{4}e^{2x}\sin(4x) ]
[ \int e^{2x} \cdot \sin(4x)dx = \frac{-1}{6}e^{2x}\cos(4x) + \frac{1}{6}e^{2x}\sin(4x) + C ]
where ( C ) is the constant of integration.
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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