int_0^{pi /2} frac{cos x}{1 + cos x + sin x} | vedclass

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(C) Let $I = \int_0^{\pi /2} \frac{\cos x}{1 + \cos x + \sin x} \,dx$. Using the property $\int_0^a f(x) \,dx = \int_0^a f(a-x) \,dx$,we get: $I = \in

Vedclass · Accepted Answer

$\frac{\pi }{4} - \frac{1}{2}\log 2$

Vedclass · Answer

(C) Let $I = \int_0^{\pi /2} \frac{\cos x}{1 + \cos x + \sin x} \,dx$. Using the property $\int_0^a f(x) \,dx = \int_0^a f(a-x) \,dx$,we get: $I = \int_0^{\pi /2} \frac{\cos(\pi/2 - x)}{1 + \cos(\pi/2 - x) + \sin(\pi/2 - x)} \,dx = \int_0^{\pi /2} \frac{\sin x}{1 + \sin x + \cos x} \,dx$. Adding the two expressions for $I$: $2I = \int_0^{\pi /2} \frac{\cos x + \sin x}{1 + \cos x + \sin x} \,dx$. Let $f(x) = 1 + \cos x + \sin x$,then $f'(x) = \cos x - \sin x$. This does not match the numerator directly. Alternatively,$2I = \int_0^{\pi /2} \frac{(1 + \cos x + \sin x) - 1}{1 + \cos x + \sin x} \,dx = \int_0^{\pi /2} 1 \,dx - \int_0^{\pi /2} \frac{1}{1 + \cos x + \sin x} \,dx$. $2I = \frac{\pi}{2} - \int_0^{\pi /2} \frac{1}{2\cos^2(x/2) + 2\sin(x/2)\cos(x/2)} \,dx = \frac{\pi}{2} - \frac{1}{2} \int_0^{\pi /2} \sec^2(x/2) \frac{1}{1 + 	an(x/2)} \,dx$. Let $u = 	an(x/2)$,then $du = \frac{1}{2} \sec^2(x/2) \,dx$. Limits: $x=0 \implies u=0$,$x=\pi/2 \implies u=1$. $2I = \frac{\pi}{2} - \int_0^1 \frac{1}{1+u} \,du = \frac{\pi}{2} - [\ln(1+u)]_0^1 = \frac{\pi}{2} - \ln 2$. $I = \frac{\pi}{4} - \frac{1}{2} \ln 2$.

$\int_0^{\pi /2} \frac{\cos x}{1 + \cos x + \sin x} \,dx = $

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