int_0^{pi / 4} log (1+tan x) d x= | vedclass

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(C) माना $I = \int_0^{\pi / 4} \log (1+	an x) d x$.गुणधर्म $\int_0^a f(x) dx = \int_0^a f(a-x) dx$ का उपयोग करने पर:$I = \int_0^{\pi / 4} \log \left[

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) माना $I = \int_0^{\pi / 4} \log (1+	an x) d x$.गुणधर्म $\int_0^a f(x) dx = \int_0^a f(a-x) dx$ का उपयोग करने पर:$I = \int_0^{\pi / 4} \log \left[1+	an \left(\frac{\pi}{4}-xight)ight] d x$.चूंकि $	an(\frac{\pi}{4}-x) = \frac{1-	an x}{1+	an x}$,समाकलन इस प्रकार होगा:$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{1+	an x + 1 - 	an x}{1+	an x}ight) d x$.$I = \int_0^{\pi / 4} \log \left(\frac{2}{1+	an x}ight) d x$.गुणधर्म $\log(\frac{a}{b}) = \log a - \log b$ का उपयोग करने पर:$I = \int_0^{\pi / 4} \log 2 d x - \int_0^{\pi / 4} \log (1+	an x) d x$.$I = \int_0^{\pi / 4} \log 2 d x - I$.$2I = \log 2 \int_0^{\pi / 4} d x = \log 2 [x]_0^{\pi / 4} = \frac{\pi}{4} \log 2$.अतः,$I = \frac{\pi}{8} \log 2$.

List-$I$	List-$II$
$I. \int_{-1}^1 x\|x\| dx$	$(a) \frac{\pi}{2}$
$II. \int_0^{\pi/2} \left(1 + \log \left(\frac{4+3\sin x}{4+3\cos x}\right)\right) dx$	$(b) \int_0^a 2f(x) dx$
$III. \int_0^a f(x) dx$	$(c) \int_0^a [f(x) + f(-x)] dx$
$IV. \int_{-a}^a f(x) dx$	$(d) 0$
	$(e) \int_0^a f(a-x) dx$

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

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