int_{pi / 4}^{pi / 3} frac{cos x-sin x}{sin 2 | vedclass

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Question

(A) Let $I = \int_{\pi / 4}^{\pi / 3} \frac{\cos x-\sin x}{\sin 2 x} d x$. We know that $\sin 2x = 1 - (1 - \sin 2x) = 1 - (\sin^2 x + \cos^2 x - 2 \s

Vedclass · Accepted Answer

$\frac{1}{2} \log \left[\frac{(3+2 \sqrt{2})(2-\sqrt{3})}{\sqrt{3}}\right]$

Vedclass · Answer

(A) Let $I = \int_{\pi / 4}^{\pi / 3} \frac{\cos x-\sin x}{\sin 2 x} d x$. We know that $\sin 2x = 1 - (1 - \sin 2x) = 1 - (\sin^2 x + \cos^2 x - 2 \sin x \cos x) = 1 - (\sin x - \cos x)^2$. Let $u = \sin x - \cos x$. Then $du = (\cos x + \sin x) dx$. This does not simplify directly. Alternatively,split the integral: $\int \frac{\cos x}{\sin 2x} dx - \int \frac{\sin x}{\sin 2x} dx = \int \frac{\cos x}{2 \sin x \cos x} dx - \int \frac{\sin x}{2 \sin x \cos x} dx = \frac{1}{2} \int \csc x dx - \frac{1}{2} \int \sec x dx$. $I = \frac{1}{2} [\log |	an(x/2)| - \log |\sec x + 	an x|]_{\pi/4}^{\pi/3} = \frac{1}{2} [\log |\frac{	an(x/2)}{\sec x + 	an x}|]_{\pi/4}^{\pi/3}$. At $x = \pi/3$: $	an(\pi/6) = 1/\sqrt{3}$,$\sec(\pi/3) = 2$,$	an(\pi/3) = \sqrt{3}$. Value $= \frac{1/\sqrt{3}}{2+\sqrt{3}} = \frac{1}{\sqrt{3}(2+\sqrt{3})} = \frac{2-\sqrt{3}}{\sqrt{3}}$. At $x = \pi/4$: $	an(\pi/8) = \sqrt{2}-1$,$\sec(\pi/4) = \sqrt{2}$,$	an(\pi/4) = 1$. Value $= \frac{\sqrt{2}-1}{\sqrt{2}+1} = (\sqrt{2}-1)^2 = 3-2\sqrt{2}$. Thus,$I = \frac{1}{2} \log \left( \frac{(2-\sqrt{3})/\sqrt{3}}{3-2\sqrt{2}} ight) = \frac{1}{2} \log \left( \frac{(2-\sqrt{3})(3+2\sqrt{2})}{\sqrt{3}} ight)$. This matches option $A$.

$\int_{\pi / 4}^{\pi / 3} \frac{\cos x-\sin x}{\sin 2 x} d x=$

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