જો I = int_0^{frac{pi}{4}} log (1 + tan x) , | vedclass

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(C) ધારો કે $I = \int_0^{\frac{\pi}{4}} \log (1 + 	an x) \, dx$. ગુણધર્મ $\int_0^a f(x) \, dx = \int_0^a f(a - x) \, dx$ નો ઉપયોગ કરતા: $I = \int_0^{

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) ધારો કે $I = \int_0^{\frac{\pi}{4}} \log (1 + 	an x) \, dx$. ગુણધર્મ $\int_0^a f(x) \, dx = \int_0^a f(a - x) \, dx$ નો ઉપયોગ કરતા: $I = \int_0^{\frac{\pi}{4}} \log \left[ 1 + 	an \left( \frac{\pi}{4} - x ight) ight] \, dx$ કારણ કે $	an \left( \frac{\pi}{4} - x ight) = \frac{1 - 	an x}{1 + 	an x}$,તેથી: $I = \int_0^{\frac{\pi}{4}} \log \left( 1 + \frac{1 - 	an x}{1 + 	an x} ight) \, dx$ $I = \int_0^{\frac{\pi}{4}} \log \left( \frac{2}{1 + 	an x} ight) \, dx$ $I = \int_0^{\frac{\pi}{4}} (\log 2 - \log (1 + 	an x)) \, dx$ $I = \int_0^{\frac{\pi}{4}} \log 2 \, dx - I$ $2I = \log 2 [x]_0^{\frac{\pi}{4}} = \frac{\pi}{4} \log 2$ $I = \frac{\pi}{8} \log 2$

જો $I = \int_0^{\frac{\pi}{4}} \log (1 + \tan x) \, dx$ હોય,તો $I$ ની કિંમત શોધો.

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