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The value of $\int_{\pi / 4}^{\pi / 2} e^{x}(\log \sin x+\cot x) d x$ is
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Verified Answer
The correct answer is:
$\frac{1}{2} e^{\pi / 4} \log 2$
Let
$\begin{array}{rl}I=\int_{\pi / 4}^{\pi / 2} e^{x}(\log \sin x & +\cot x) d x \\ \Rightarrow I=\int_{\pi / 4}^{\pi / 2} e^{x} \log \sin x & d x \\ +\int_{\pi / 4}^{\pi / 2} e^{x} \cot x d x \\ =\int_{\pi / 4}^{\pi / 2} e^{x} \log \sin x d x+\left[e^{x} \log \sin x\right]_{\pi / 4}^{\pi / 2} \\ -\int_{\pi / 4}^{\pi / 2} e^{x} \log \sin x d x \\ = & e^{\pi / 2} \log \sin \frac{\pi}{2}-e^{\pi / 4} \log \sin \frac{\pi}{4} \\ = & -e^{\pi / 4} \log \left(\frac{1}{\sqrt{2}}\right) \\ = & \frac{1}{2} e^{\pi / 4} \log 2\end{array}$
$\begin{array}{rl}I=\int_{\pi / 4}^{\pi / 2} e^{x}(\log \sin x & +\cot x) d x \\ \Rightarrow I=\int_{\pi / 4}^{\pi / 2} e^{x} \log \sin x & d x \\ +\int_{\pi / 4}^{\pi / 2} e^{x} \cot x d x \\ =\int_{\pi / 4}^{\pi / 2} e^{x} \log \sin x d x+\left[e^{x} \log \sin x\right]_{\pi / 4}^{\pi / 2} \\ -\int_{\pi / 4}^{\pi / 2} e^{x} \log \sin x d x \\ = & e^{\pi / 2} \log \sin \frac{\pi}{2}-e^{\pi / 4} \log \sin \frac{\pi}{4} \\ = & -e^{\pi / 4} \log \left(\frac{1}{\sqrt{2}}\right) \\ = & \frac{1}{2} e^{\pi / 4} \log 2\end{array}$
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