By using the properties of definite integrals | vedclass

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(A) Let $I = \int_{0}^{\frac{\pi}{4}} \log (1+	an x) d x$ ..... $(1)$Using the property $\int_{0}^{a} f(x) d x = \int_{0}^{a} f(a-x) d x$,we get:$I =

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

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(A) Let $I = \int_{0}^{\frac{\pi}{4}} \log (1+	an x) d x$ ..... $(1)$Using the property $\int_{0}^{a} f(x) d x = \int_{0}^{a} f(a-x) d x$,we get:$I = \int_{0}^{\frac{\pi}{4}} \log \left[1+	an \left(\frac{\pi}{4}-xight)ight] d x$Using $	an(A-B) = \frac{	an A - 	an B}{1 + 	an A 	an B}$:$I = \int_{0}^{\frac{\pi}{4}} \log \left\{1+\frac{	an \frac{\pi}{4}-	an x}{1+	an \frac{\pi}{4} 	an x}ight\} d x$Since $	an \frac{\pi}{4} = 1$:$I = \int_{0}^{\frac{\pi}{4}} \log \left\{1+\frac{1-	an x}{1+	an x}ight\} d x$$I = \int_{0}^{\frac{\pi}{4}} \log \left\{\frac{1+	an x + 1 - 	an x}{1+	an x}ight\} d x$$I = \int_{0}^{\frac{\pi}{4}} \log \left(\frac{2}{1+	an x}ight) d x$Using $\log(\frac{a}{b}) = \log a - \log b$:$I = \int_{0}^{\frac{\pi}{4}} \log 2 d x - \int_{0}^{\frac{\pi}{4}} \log (1+	an x) d x$$I = \int_{0}^{\frac{\pi}{4}} \log 2 d x - I$ (From equation $(1)$)$2I = [x \log 2]_{0}^{\frac{\pi}{4}}$$2I = \frac{\pi}{4} \log 2$$I = \frac{\pi}{8} \log 2$

By using the properties of definite integrals,evaluate the integral $\int_{0}^{\frac{\pi}{4}} \log (1+\tan x) d x$.

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