The value of the integral int_{0}^{infty} e^{ | vedclass

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(C) Let $I = \int_{0}^{\infty} e^{-2x} (\sin 2x + \cos 2x) dx$.We use the standard integral formula: $\int e^{ax} (f(x) + \frac{1}{a} f'(x)) dx = \fra

Vedclass · Accepted Answer

$1/2$

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(C) Let $I = \int_{0}^{\infty} e^{-2x} (\sin 2x + \cos 2x) dx$.We use the standard integral formula: $\int e^{ax} (f(x) + \frac{1}{a} f'(x)) dx = \frac{1}{a} e^{ax} f(x) + C$.Here,$a = -2$ and $f(x) = \sin 2x$.Then $f'(x) = 2 \cos 2x$.Note that $\sin 2x + \cos 2x = \sin 2x + \frac{1}{-2} (-2 \cos 2x) = f(x) + \frac{1}{a} f'(x)$.Thus,the integral is $\left[ \frac{1}{-2} e^{-2x} \sin 2x ight]_{0}^{\infty}$.Evaluating at the limits:At $x 	o \infty$,$e^{-2x} 	o 0$,so the term becomes $0$.At $x = 0$,$\sin(0) = 0$,so the term becomes $0$.However,let us evaluate the integral directly using $\int e^{ax} \sin bx dx = \frac{e^{ax}}{a^2+b^2} (a \sin bx - b \cos bx)$ and $\int e^{ax} \cos bx dx = \frac{e^{ax}}{a^2+b^2} (a \cos bx + b \sin bx)$.For $I = \int_{0}^{\infty} e^{-2x} \sin 2x dx + \int_{0}^{\infty} e^{-2x} \cos 2x dx$:$\int_{0}^{\infty} e^{-2x} \sin 2x dx = \left[ \frac{e^{-2x}}{(-2)^2 + 2^2} (-2 \sin 2x - 2 \cos 2x) ight]_{0}^{\infty} = \left[ \frac{e^{-2x}}{8} (-2 \sin 2x - 2 \cos 2x) ight]_{0}^{\infty} = 0 - (\frac{1}{8} (-2)) = 1/4$.$\int_{0}^{\infty} e^{-2x} \cos 2x dx = \left[ \frac{e^{-2x}}{(-2)^2 + 2^2} (-2 \cos 2x + 2 \sin 2x) ight]_{0}^{\infty} = \left[ \frac{e^{-2x}}{8} (-2 \cos 2x + 2 \sin 2x) ight]_{0}^{\infty} = 0 - (\frac{1}{8} (-2)) = 1/4$.Adding these,$I = 1/4 + 1/4 = 1/2$.

The value of the integral $\int_{0}^{\infty} e^{-2x} (\sin 2x + \cos 2x) dx$ is:

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