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

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(D) ધારો કે $I = \int {e^{x/2}} \sin \left( \frac{x}{2} + \frac{\pi}{4} \right) \, dx$. સૂત્ર $\int e^{ax} \sin(bx + c) \, dx = \frac{e^{ax}}{a^2 + b^

Vedclass · Accepted Answer

$\sqrt{2} {e^{x/2}} \sin \frac{x}{2} + c$

Vedclass · Answer

(D) ધારો કે $I = \int {e^{x/2}} \sin \left( \frac{x}{2} + \frac{\pi}{4} \right) \, dx$. સૂત્ર $\int e^{ax} \sin(bx + c) \, dx = \frac{e^{ax}}{a^2 + b^2} [a \sin(bx + c) - b \cos(bx + c)] + C$ નો ઉપયોગ કરતા. અહીં $a = 1/2$ અને $b = 1/2$ છે. $I = \frac{e^{x/2}}{(1/2)^2 + (1/2)^2} \left[ \frac{1}{2} \sin \left( \frac{x}{2} + \frac{\pi}{4} \right) - \frac{1}{2} \cos \left( \frac{x}{2} + \frac{\pi}{4} \right) \right] + C$. $I = \frac{e^{x/2}}{1/4 + 1/4} \cdot \frac{1}{2} \left[ \sin \left( \frac{x}{2} + \frac{\pi}{4} \right) - \cos \left( \frac{x}{2} + \frac{\pi}{4} \right) \right] + C$. $I = \frac{e^{x/2}}{1/2} \cdot \frac{1}{2} \left[ \sin \left( \frac{x}{2} + \frac{\pi}{4} \right) - \cos \left( \frac{x}{2} + \frac{\pi}{4} \right) \right] + C$. $I = e^{x/2} \left[ \sin \frac{x}{2} \cos \frac{\pi}{4} + \cos \frac{x}{2} \sin \frac{\pi}{4} - (\cos \frac{x}{2} \cos \frac{\pi}{4} - \sin \frac{x}{2} \sin \frac{\pi}{4}) \right] + C$. કેમ કે $\sin \frac{\pi}{4} = \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$,$I = e^{x/2} \left[ \frac{1}{\sqrt{2}} \sin \frac{x}{2} + \frac{1}{\sqrt{2}} \cos \frac{x}{2} - \frac{1}{\sqrt{2}} \cos \frac{x}{2} + \frac{1}{\sqrt{2}} \sin \frac{x}{2} \right] + C$. $I = e^{x/2} \left[ \frac{2}{\sqrt{2}} \sin \frac{x}{2} \right] + C = \sqrt{2} {e^{x/2}} \sin \frac{x}{2} + C$. આમ,સાચો વિકલ્પ $D$ છે.

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

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