The value of int_{pi / 4}^{pi / 2} e^{x}(log | vedclass

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(C) We use the standard result $\int e^x (f(x) + f'(x)) dx = e^x f(x) + C$. Let $f(x) = \log \sin x$. Then $f'(x) = \frac{1}{\sin x} \cdot \cos x = \c

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

Vedclass · Answer

(C) We use the standard result $\int e^x (f(x) + f'(x)) dx = e^x f(x) + C$. Let $f(x) = \log \sin x$. Then $f'(x) = \frac{1}{\sin x} \cdot \cos x = \cot x$. Thus,the integral becomes $\int_{\pi / 4}^{\pi / 2} e^x (f(x) + f'(x)) dx = [e^x f(x)]_{\pi / 4}^{\pi / 2}$. Substituting the limits: $= [e^x \log \sin x]_{\pi / 4}^{\pi / 2} = e^{\pi / 2} \log \sin(\pi / 2) - e^{\pi / 4} \log \sin(\pi / 4)$. Since $\sin(\pi / 2) = 1$ and $\log(1) = 0$,the first term is $0$. $= 0 - e^{\pi / 4} \log(1 / \sqrt{2}) = -e^{\pi / 4} \log(2^{-1/2})$. $= -e^{\pi / 4} \cdot (-1/2) \log 2 = \frac{1}{2} e^{\pi / 4} \log 2$.

The value of $\int_{\pi / 4}^{\pi / 2} e^{x}(\log \sin x+\cot x) d x$ is

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