To solve the integral ∫02πe2xcos(2x)dx, we will use integration by parts.
Recall that integration by parts is given by:
∫udv=uv−∫vdu
We need to choose u and dv. Let's set:
u=e2xanddv=cos(2x)dx
Then we need to find du and v:
du=2e2xdxandv=21sin(2x)
Now we apply integration by parts:
∫e2xcos(2x)dx=e2x⋅21sin(2x)−∫21sin(2x)⋅2e2xdx=21e2xsin(2x)−∫e2xsin(2x)dx
Now we need to integrate ∫e2xsin(2x)dx. We use integration by parts again for this integral. Set:
u=e2xanddv=sin(2x)dx
Then:
du=2e2xdxandv=−21cos(2x)
Now applying integration by parts again:
∫e2xsin(2x)dx=−21e2xcos(2x)−∫−21cos(2x)⋅2e2xdx=−21e2xcos(2x)+∫e2xcos(2x)dx
Notice that we end up with the original integral ∫e2xcos(2x)dx on the right-hand side. Let's call this integral I:
I=∫e2xcos(2x)dx
So we have:
I=21e2xsin(2x)−(−21e2xcos(2x))−II=21e2xsin(2x)+21e2xcos(2x)−I2I=21e2xsin(2x)+21e2xcos(2x)I=41e2x(sin(2x)+cos(2x))
Now we need to evaluate this expression from 0 to 2π:
41e2x(sin(2x)+cos(2x))02π
Evaluating at the bounds:
41e2x(sin(2x)+cos(2x))02π=41e2⋅2π(sin(π)+cos(π))−41e2⋅0(sin(0)+cos(0))=41eπ(0+(−1))−41(0+1)=−41eπ−41=−41(eπ+1)
So the value of the integral is:
−41(eπ+1)
Would you like more details or have any questions?
Here are some related questions:
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What are some applications of definite integrals?
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Tip: When using integration by parts, choosing u and dv wisely can simplify the process significantly.
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Math Problem Analysis
Mathematical Concepts
Integration Exponential Functions Trigonometric Functions Integration by Parts