int e^x cdot cos 2x , dx = . . . . . . + C | vedclass

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(D) Let $I = \int e^x \cos 2x \, dx$. Using the integration by parts formula $\int u \cdot v \, dx = u \int v \, dx - \int (u' \int v \, dx) \, dx$,le

Vedclass · Accepted Answer

$\frac{e^x}{5}(\cos 2x + 2 \sin 2x)$

Vedclass · Answer

(D) Let $I = \int e^x \cos 2x \, dx$. Using the integration by parts formula $\int u \cdot v \, dx = u \int v \, dx - \int (u' \int v \, dx) \, dx$,let $u = \cos 2x$ and $v = e^x$. $I = \cos 2x \cdot e^x - \int (-2 \sin 2x) \cdot e^x \, dx = e^x \cos 2x + 2 \int e^x \sin 2x \, dx$. Applying integration by parts again to $\int e^x \sin 2x \, dx$ with $u = \sin 2x$ and $v = e^x$: $\int e^x \sin 2x \, dx = \sin 2x \cdot e^x - \int (2 \cos 2x) \cdot e^x \, dx = e^x \sin 2x - 2I$. Substituting this back into the equation for $I$: $I = e^x \cos 2x + 2(e^x \sin 2x - 2I) = e^x \cos 2x + 2e^x \sin 2x - 4I$. $5I = e^x(\cos 2x + 2 \sin 2x)$. $I = \frac{e^x}{5}(\cos 2x + 2 \sin 2x) + C$. Thus,the correct option is $D$.

$\int e^x \cdot \cos 2x \, dx = $ . . . . . . $+ C$.

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