int_0^{2pi} e^{x/2} sin left( frac{x}{2} + fr | vedclass

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(C) माना $I = \int_0^{2\pi} e^{x/2} \sin \left( \frac{x}{2} + \frac{\pi}{4} \right) \, dx$. $t = \frac{x}{2}$ प्रतिस्थापित करने पर,$dt = \frac{1}{2} d

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(C) माना $I = \int_0^{2\pi} e^{x/2} \sin \left( \frac{x}{2} + \frac{\pi}{4} \right) \, dx$. $t = \frac{x}{2}$ प्रतिस्थापित करने पर,$dt = \frac{1}{2} dx$ प्राप्त होता है,जिसका अर्थ है $dx = 2 dt$. जब $x = 0$,तब $t = 0$. जब $x = 2\pi$,तब $t = \pi$. अतः,$I = \int_0^{\pi} e^t \sin \left( t + \frac{\pi}{4} \right) (2 dt) = 2 \int_0^{\pi} e^t \sin \left( t + \frac{\pi}{4} \right) dt$. मानक समाकलन सूत्र $\int e^{at} \sin(bt + c) dt = \frac{e^{at}}{a^2 + b^2} [a \sin(bt + c) - b \cos(bt + c)]$ का उपयोग करने पर,जहाँ $a=1, b=1, c=\frac{\pi}{4}$: $I = 2 \left[ \frac{e^t}{1^2 + 1^2} (\sin(t + \frac{\pi}{4}) - \cos(t + \frac{\pi}{4})) \right]_0^{\pi}$. $I = 2 \left[ \frac{e^t}{2} (\sin(t + \frac{\pi}{4}) - \cos(t + \frac{\pi}{4})) \right]_0^{\pi} = [e^t (\sin(t + \frac{\pi}{4}) - \cos(t + \frac{\pi}{4}))]_0^{\pi}$. $t = \pi$ पर: $e^{\pi} (\sin(\pi + \frac{\pi}{4}) - \cos(\pi + \frac{\pi}{4})) = e^{\pi} (-\sin \frac{\pi}{4} - (-\cos \frac{\pi}{4})) = e^{\pi} (-\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) = 0$. $t = 0$ पर: $e^0 (\sin \frac{\pi}{4} - \cos \frac{\pi}{4}) = 1 (\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}) = 0$. अतः,$I = 0 - 0 = 0$.

$\int_0^{2\pi} e^{x/2} \sin \left( \frac{x}{2} + \frac{\pi}{4} \right) \, dx = $

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