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tangare [24]
3 years ago
5

Integrate e^x(sin(x) cos(x))

Mathematics
1 answer:
Karo-lina-s [1.5K]3 years ago
6 0
I=\int e^x(\sin(x)\cos(x))dx=\int e^x(\frac{1}{2}\sin(2x))dx=\frac{1}{2}\int e^x\sin(2x)dx

\text{If }u=\sin(2x)\to du=2\cos(2x)dx~\text{and}~dv=e^xdx\to v=e^x:\\\\
\text{Using }\int u\,dv=uv-\int v\,du:\\\\
I=\frac{1}{2}(e^x\sin(2x)-\int e^x(2\cos(2x))dx)\\\\
2I=e^x\sin(2x)-2\underbrace{\int e^x\cos(2x)dx}_{I_2}

Looking for I_2:

\text{If}~u=\cos(2x)\to du=-2\sin(2x)dx~\text{and}~dv=e^xdx\to v=e^x:\\\\
I_2=e^x\cos(2x)-\int e^x(-2\sin(2x))dx\\\\ I_2=e^x\cos(2x)+2\int e^x(\sin(2x))dx\\\\  I_2=e^x\cos(2x)+2\int e^x(2\sin(x)\cos(x))dx\\\\ I_2=e^x\cos(2x)+4\int e^x(\sin(x)\cos(x))dx=e^x\cos(2x)+4I

Replacing:

2I=e^x\sin(2x)-2I_2\iff\\\\2I=e^x\sin(2x)-2(e^x\cos(2x)+4I)\iff\\\\
2I=e^x\sin(2x)-2e^x\cos(2x)-8I\iff\\\\
10I=e^x\sin(2x)-2e^x\cos(2x)\iff\\\\
I=\dfrac{e^x}{10}(\sin(2x)-2\cos(2x))\\\\
\boxed{\int e^x(\sin(x)\cos(x))dx=\dfrac{e^x}{10}(\sin(2x)-2\cos(2x))+C}
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Step-by-step explanation:

Four less ( minus 4 ) than the product of ten and a number ( 10 times x ) is eleven ( equal to 11 ).

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Step-by-step explanation:

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Check that 2b+b and 3b have the same value when b is 1,2, and 3. Do 2b+b and 3b have the same value for all values of b? Explain
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Answer:

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Step-by-step explanation:

Given the two expression 2b+b and 3b and values of b as 1, 2 and 3.

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Substituting the value of b=2,

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Substituting the value of b=3,

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Substituting the value of b=1,

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Substituting the value of b=2,

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Substituting the value of b=3,

3\left(3\right)=9 ....6

For value of b=1, equation 1 and equation 4 are same, for value of b=2, equation 2 and equation 5 are same and for value of b=3 equation 3 and equation 6 are same.

Therefore both expression are same for all values of b.

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