int_0^infty {{e^{ - 2x}}(sin 2x + cos 2x),dx | vedclass

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Question

(C) Let $I = \int_0^\infty e^{-2x} (\sin 2x + \cos 2x) \, dx$. Using the standard integral formula $\int e^{ax} \sin(bx) \, dx = \frac{e^{ax}}{a^2 + b

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(C) Let $I = \int_0^\infty e^{-2x} (\sin 2x + \cos 2x) \, dx$. Using the standard integral formula $\int e^{ax} \sin(bx) \, dx = \frac{e^{ax}}{a^2 + b^2} (a \sin bx - b \cos bx)$ and $\int e^{ax} \cos(bx) \, dx = \frac{e^{ax}}{a^2 + b^2} (a \cos bx + b \sin bx)$. Here $a = -2$ and $b = 2$. $\int_0^\infty e^{-2x} \sin 2x \, dx = \left[ \frac{e^{-2x}}{(-2)^2 + 2^2} (-2 \sin 2x - 2 \cos 2x) \right]_0^\infty = \left[ \frac{e^{-2x}}{8} (-2 \sin 2x - 2 \cos 2x) \right]_0^\infty = 0 - (\frac{1}{8}(-2)) = \frac{1}{4}$. $\int_0^\infty e^{-2x} \cos 2x \, dx = \left[ \frac{e^{-2x}}{(-2)^2 + 2^2} (-2 \cos 2x + 2 \sin 2x) \right]_0^\infty = \left[ \frac{e^{-2x}}{8} (-2 \cos 2x + 2 \sin 2x) \right]_0^\infty = 0 - (\frac{1}{8}(-2)) = \frac{1}{4}$. Adding both results: $I = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$.

$\int_0^\infty {{e^{ - 2x}}(\sin 2x + \cos 2x)\,dx = } $

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