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प्रश्न
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उत्तर
\[Let\ I = \int_0^{2\pi} e^\frac{x}{2} \sin \left( \frac{x}{2} + \frac{\pi}{4} \right) d x . Then, \]
\[\text{Integrating by parts}\]
\[I = \left[ - 2 e^\frac{x}{2} \cos \left( \frac{x}{2} + \frac{\pi}{4} \right) \right]_0^{2\pi} - \int_0^{2\pi} - \frac{2}{2} e^\frac{x}{2} \cos \left( \frac{x}{2} + \frac{\pi}{4} \right) dx\]
\[\text{Again, integrating second term by parts}\]
\[ \Rightarrow I = \left[ - 2 e^\frac{x}{2} \cos \left( \frac{x}{2} + \frac{\pi}{4} \right) \right]_0^{2\pi} + \left\{ \left[ 2 e^\frac{x}{2} \sin \left( \frac{x}{2} + \frac{\pi}{4} \right) \right]_0^{2\pi} - \int_0^{2\pi} \frac{2}{2} e^\frac{x}{2} \sin \left( \frac{x}{2} + \frac{\pi}{4} \right) d x \right\}\]
\[ \Rightarrow I = \left[ - 2 e^\frac{x}{2} \cos \left( \frac{x}{2} + \frac{\pi}{4} \right) \right]_0^{2\pi} + \left[ 2 e^\frac{x}{2} \sin \left( \frac{x}{2} + \frac{\pi}{4} \right) \right]_0^{2\pi} - I\]
\[ \Rightarrow 2I = \frac{2}{\sqrt{2}} e^\pi + \frac{2}{\sqrt{2}} - \frac{2}{\sqrt{2}} e^\pi - \frac{2}{\sqrt{2}} = 0\]
\[ \Rightarrow I = 0\]
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