tan ^{-1}left[int_{-pi / 2}^{pi / 2} frac{cos | vedclass

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Question

(A) Let $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\cos x}{1+e^x} dx$ ... $(i)$ Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x)

Vedclass · Accepted Answer

$\frac{\pi}{4}$

Vedclass · Answer

(A) Let $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\cos x}{1+e^x} dx$ ... $(i)$ Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,where $a = -\frac{\pi}{2}$ and $b = \frac{\pi}{2}$,we have $a+b = 0$. $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\cos(-x)}{1+e^{-x}} dx = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\cos x}{1+\frac{1}{e^x}} dx = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{e^x \cos x}{e^x+1} dx$ ... (ii) Adding $(i)$ and (ii): $2I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\cos x + e^x \cos x}{1+e^x} dx = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\cos x(1+e^x)}{1+e^x} dx = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos x dx$ $2I = [\sin x]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = \sin(\frac{\pi}{2}) - \sin(-\frac{\pi}{2}) = 1 - (-1) = 2$ $I = 1$ Therefore,$	an^{-1}(I) = 	an^{-1}(1) = \frac{\pi}{4}$.

$\tan ^{-1}\left[\int_{-\pi / 2}^{\pi / 2} \frac{\cos x}{1+e^x} d x\right]=$

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