int_{pi/4}^{pi/2} e^x (log sin x + cot x) , d | vedclass

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(C) Let $I = \int_{\pi/4}^{\pi/2} e^x (\log \sin x + \cot x) \, dx$. Using the property $\int e^x (f(x) + f'(x)) \, dx = e^x f(x) + C$,we identify $f(

Vedclass · Accepted Answer

$\frac{1}{2} e^{\pi/4} \log 2$

Vedclass · Answer

(C) Let $I = \int_{\pi/4}^{\pi/2} e^x (\log \sin x + \cot x) \, dx$. Using the property $\int e^x (f(x) + f'(x)) \, dx = e^x f(x) + C$,we identify $f(x) = \log \sin x$. Then $f'(x) = \frac{1}{\sin x} \cdot \cos x = \cot x$. Thus,the integral becomes: $I = [e^x \log \sin x]_{\pi/4}^{\pi/2}$ $I = e^{\pi/2} \log \sin(\pi/2) - e^{\pi/4} \log \sin(\pi/4)$ Since $\sin(\pi/2) = 1$ and $\log(1) = 0$: $I = e^{\pi/2} \cdot 0 - e^{\pi/4} \log(1/\sqrt{2})$ $I = -e^{\pi/4} \log(2^{-1/2}) = -e^{\pi/4} \cdot (-\frac{1}{2} \log 2) = \frac{1}{2} e^{\pi/4} \log 2$.

$\int_{\pi/4}^{\pi/2} e^x (\log \sin x + \cot x) \, dx = $

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