int_0^{pi /4} {log (1 + tan theta ),dtheta = | vedclass

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(C) Let $I = \int_0^{\pi /4} \log (1 + 	an 	heta )\,d	heta$  Using the property $\int_0^a f(x)dx = \int_0^a f(a-x)dx$,we get:  $I = \int_0^{\pi /4}

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

Vedclass · Answer

(C) Let $I = \int_0^{\pi /4} \log (1 + 	an 	heta )\,d	heta$  Using the property $\int_0^a f(x)dx = \int_0^a f(a-x)dx$,we get:  $I = \int_0^{\pi /4} \log \left( 1 + 	an \left( \frac{\pi }{4} - 	heta ight) ight) d	heta$  Since $	an(A-B) = \frac{	an A - 	an B}{1 + 	an A 	an B}$,we have $	an(\frac{\pi}{4} - 	heta) = \frac{1 - 	an 	heta}{1 + 	an 	heta}$  $I = \int_0^{\pi /4} \log \left( 1 + \frac{1 - 	an 	heta}{1 + 	an 	heta} ight) d	heta$  $I = \int_0^{\pi /4} \log \left( \frac{1 + 	an 	heta + 1 - 	an 	heta}{1 + 	an 	heta} ight) d	heta$  $I = \int_0^{\pi /4} \log \left( \frac{2}{1 + 	an 	heta} ight) d	heta$  $I = \int_0^{\pi /4} (\log 2 - \log (1 + 	an 	heta )) d	heta$  $I = \int_0^{\pi /4} \log 2 \, d	heta - \int_0^{\pi /4} \log (1 + 	an 	heta ) d	heta$  $I = \log 2 [	heta]_0^{\pi /4} - I$  $2I = \frac{\pi}{4} \log 2$  $I = \frac{\pi}{8} \log 2$

$\int_0^{\pi /4} {\log (1 + \tan \theta )\,d\theta = } $

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