int_{0}^{pi / 4} log left(frac{sin x+cos x}{ | vedclass

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(C) Let $I = \int_{0}^{\pi / 4} \log \left(\frac{\sin x+\cos x}{\cos x}\right) d x$. Simplifying the integrand,we get $I = \int_{0}^{\pi / 4} \log (1+

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$ \frac{\pi}{8} \log 2 $

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(C) Let $I = \int_{0}^{\pi / 4} \log \left(\frac{\sin x+\cos x}{\cos x}ight) d x$. Simplifying the integrand,we get $I = \int_{0}^{\pi / 4} \log (1+	an x) d x$. Using the property $\int_{0}^{a} f(x) d x = \int_{0}^{a} f(a-x) d x$,we have: $I = \int_{0}^{\pi / 4} \log \left(1+	an \left(\frac{\pi}{4}-xight)ight) d x$. Since $	an(\frac{\pi}{4}-x) = \frac{1-	an x}{1+	an x}$,the integral becomes: $I = \int_{0}^{\pi / 4} \log \left(1+\frac{1-	an x}{1+	an x}ight) d x = \int_{0}^{\pi / 4} \log \left(\frac{2}{1+	an x}ight) d x$. Using the property $\log(\frac{a}{b}) = \log a - \log b$: $I = \int_{0}^{\pi / 4} \log 2 d x - \int_{0}^{\pi / 4} \log (1+	an x) d x$. $I = \log 2 [x]_{0}^{\pi / 4} - I$. $2I = \frac{\pi}{4} \log 2$. Therefore,$I = \frac{\pi}{8} \log 2$.

$ \int_{0}^{\pi / 4} \log \left(\frac{\sin x+\cos x}{\cos x}\right) d x $

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