माना I=int_{-frac{pi}{4}}^{frac{pi}{4}} frac{ | vedclass

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(A) माना $I=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{1}{2-\cos 2 x}\left(\frac{3}{\pi}+\log \left(\frac{4+\sin x}{4-\sin x}\right)\right) d x$. समा

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$4$

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(A) माना $I=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{1}{2-\cos 2 x}\left(\frac{3}{\pi}+\log \left(\frac{4+\sin x}{4-\sin x}ight)ight) d x$. समाकलन को दो भागों में विभाजित करें: $I=\frac{3}{\pi} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{1}{2-\cos 2 x} d x + \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{1}{2-\cos 2 x} \log \left(\frac{4+\sin x}{4-\sin x}ight) d x$. दूसरा समाकलन शून्य है क्योंकि फलन विषम है। अतः, $I=\frac{3}{\pi} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{1}{2-\cos 2 x} d x = \frac{6}{\pi} \int_0^{\frac{\pi}{4}} \frac{1}{2-\cos 2 x} d x$. $\cos 2x = \frac{1-	an^2 x}{1+	an^2 x}$ का उपयोग करते हुए: $I=\frac{6}{\pi} \int_0^{\frac{\pi}{4}} \frac{\sec^2 x}{2(1+	an^2 x) - (1-	an^2 x)} d x = \frac{6}{\pi} \int_0^{\frac{\pi}{4}} \frac{\sec^2 x}{1+3	an^2 x} d x$. माना $	an x = t$, तो $\sec^2 x d x = d t$. जब $x: 0 	o \frac{\pi}{4}$, तब $t: 0 	o 1$. $I=\frac{6}{\pi} \int_0^1 \frac{d t}{1+3t^2} = \frac{6}{\pi} \left[ \frac{1}{\sqrt{3}} 	an^{-1}(\sqrt{3}t) ight]_0^1 = \frac{6}{\pi \sqrt{3}} 	an^{-1}(\sqrt{3}) = \frac{2\sqrt{3}}{\pi} \cdot \frac{\pi}{3} = \frac{2}{\sqrt{3}}$. इसलिए, $3I^2 = 3 \left( \frac{2}{\sqrt{3}} ight)^2 = 3 \cdot \frac{4}{3} = 4$.

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

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