Question

only answer with full explanation​

Answer 1

Answer

⇒ $I = \frac{e^2^xSinx + 2e^2^xCosx}{5}$

Step-by-step explanation ⇒

I = $\int {e^xCosx} \, dx$

$I = e^2^x\int\ {Cosx} \, dx  - \int{ [\frac{de^2^x}{dx} \int {Cosx} \, dx] } \, dx\\I = e^2^xSinx - 2\int{e^2^xSinx} \, dx\\I = e^2^xSinx - 2e^2^x\int\ {Sinx} \, dx  - 2\int{ [\frac{de^2^x}{dx} \int {Sinx} \, dx] }\\I = e^2^xSinx + 2e^2^xCosx - 4\int{e^2^xCosx} \, dx\\I = \frac{e^2^xSinx + 2e^2^xCosx}{5}$

• Thus, The answer for this question is = $I = \frac{e^2^xSinx + 2e^2^xCosx}{5}$

• UV Method of integration is applied fort solution

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Answer 2

Step-by-step explanation:

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