int_0^{frac{pi}{4}} log left(frac{sin x+cos x | vedclass

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Question

(D) Let $I = \int_0^{\frac{\pi}{4}} \log \left(\frac{\sin x + \cos x}{\cos x}\right) dx$. Simplifying the integrand,we get $I = \int_0^{\frac{\pi}{4}}

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

Vedclass · Answer

(D) Let $I = \int_0^{\frac{\pi}{4}} \log \left(\frac{\sin x + \cos x}{\cos x}ight) dx$. Simplifying the integrand,we get $I = \int_0^{\frac{\pi}{4}} \log(1 + 	an x) dx$. Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we have: $I = \int_0^{\frac{\pi}{4}} \log \left[1 + 	an \left(\frac{\pi}{4} - xight)ight] dx$. Using the formula $	an(A-B) = \frac{	an A - 	an B}{1 + 	an A 	an B}$,we get: $I = \int_0^{\frac{\pi}{4}} \log \left[1 + \frac{1 - 	an x}{1 + 	an x}ight] dx$. $I = \int_0^{\frac{\pi}{4}} \log \left(\frac{1 + 	an x + 1 - 	an x}{1 + 	an x}ight) dx$. $I = \int_0^{\frac{\pi}{4}} \log \left(\frac{2}{1 + 	an x}ight) dx$. $I = \int_0^{\frac{\pi}{4}} (\log 2 - \log(1 + 	an x)) dx$. $I = \int_0^{\frac{\pi}{4}} \log 2 dx - \int_0^{\frac{\pi}{4}} \log(1 + 	an x) dx$. $I = \log 2 [x]_0^{\frac{\pi}{4}} - I$. $2I = \frac{\pi}{4} \log 2$. $I = \frac{\pi}{8} \log 2$.

$\int_0^{\frac{\pi}{4}} \log \left(\frac{\sin x+\cos x}{\cos x}\right) d x=$

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