int_{ - pi /4}^{pi /2} {{e^{ - x}}sin x,dx} = | vedclass

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Question

(A) We use the standard integral formula: $\int e^{ax} \sin(bx) dx = \frac{e^{ax}}{a^2 + b^2} (a \sin(bx) - b \cos(bx)) + C$. Here,$a = -1$ and $b = 1

Vedclass · Accepted Answer

$ - \frac{1}{2}{e^{ - \pi /2}}$

Vedclass · Answer

(A) We use the standard integral formula: $\int e^{ax} \sin(bx) dx = \frac{e^{ax}}{a^2 + b^2} (a \sin(bx) - b \cos(bx)) + C$. Here,$a = -1$ and $b = 1$. Thus,$\int e^{-x} \sin x dx = \frac{e^{-x}}{(-1)^2 + 1^2} (-1 \sin x - 1 \cos x) = -\frac{e^{-x}}{2} (\sin x + \cos x)$. Now,applying the limits from $-\pi/4$ to $\pi/2$: $= [-\frac{e^{-x}}{2} (\sin x + \cos x)]_{-\pi/4}^{\pi/2}$ $= -\frac{1}{2} [e^{-\pi/2} (\sin(\pi/2) + \cos(\pi/2)) - e^{\pi/4} (\sin(-\pi/4) + \cos(-\pi/4))]$ $= -\frac{1}{2} [e^{-\pi/2} (1 + 0) - e^{\pi/4} (-\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}})]$ $= -\frac{1}{2} [e^{-\pi/2} (1) - e^{\pi/4} (0)]$ $= -\frac{1}{2} e^{-\pi/2}$.

$\int_{ - \pi /4}^{\pi /2} {{e^{ - x}}\sin x\,dx} = $

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