int e^{x / 2}left(frac{2+sin x}{1+cos x}right | vedclass

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(B) We have the integral $I = \int e^{x / 2} \left( \frac{2 + \sin x}{1 + \cos x} ight) dx$. Using the half-angle formulas $\sin x = \frac{2 	an(x/

Vedclass · Accepted Answer

$2 e^{x / 2} 	an \left(\frac{x}{2}ight)+c$

Vedclass · Answer

(B) We have the integral $I = \int e^{x / 2} \left( \frac{2 + \sin x}{1 + \cos x} ight) dx$. Using the half-angle formulas $\sin x = \frac{2 	an(x/2)}{1 + 	an^2(x/2)}$ and $1 + \cos x = 2 \cos^2(x/2)$,we can simplify the integrand: $\frac{2 + \sin x}{1 + \cos x} = \frac{2}{2 \cos^2(x/2)} + \frac{2 	an(x/2)}{(1 + 	an^2(x/2)) \cdot 2 \cos^2(x/2)}$ $= \sec^2(x/2) + \frac{	an(x/2)}{\sec^2(x/2) \cdot \cos^2(x/2)} = \sec^2(x/2) + 	an(x/2)$. Thus,$I = \int e^{x/2} (\sec^2(x/2) + 	an(x/2)) dx$. Let $f(x) = 	an(x/2)$,then $f'(x) = \sec^2(x/2) \cdot \frac{1}{2}$. This does not directly fit the form $\int e^x (f(x) + f'(x)) dx$. Let $u = x/2$,then $du = dx/2$,so $dx = 2du$. $I = \int e^u (\sec^2 u + 	an u) \cdot 2 du = 2 \int e^u (	an u + \sec^2 u) du$. Since $\frac{d}{du}(	an u) = \sec^2 u$,using the formula $\int e^u (f(u) + f'(u)) du = e^u f(u) + c$,we get: $I = 2 e^u 	an u + c = 2 e^{x/2} 	an(x/2) + c$. Hence,option $B$ is correct.

$\int e^{x / 2}\left(\frac{2+\sin x}{1+\cos x}\right) d x=$

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