text{જો } int frac{1}{operatorname{cosec} x+c | vedclass

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(A) $	ext{આપેલ } I = \int \frac{1}{\operatorname{cosec} x+\cos x} d x = \int \frac{\sin x}{1+\sin x \cos x} d x = \int \frac{2 \sin x}{2+\sin 2 x} d

Vedclass · Accepted Answer

$\frac{3 \sqrt{3}-1}{\sqrt{3}+1}$

Vedclass · Answer

(A) $	ext{આપેલ } I = \int \frac{1}{\operatorname{cosec} x+\cos x} d x = \int \frac{\sin x}{1+\sin x \cos x} d x = \int \frac{2 \sin x}{2+\sin 2 x} d x$ $	ext{આપણે } 2 \sin x = (\sin x + \cos x) - (\cos x - \sin x) 	ext{ લખી શકીએ.}$ $	ext{તેથી, } I = \int \frac{\sin x + \cos x}{2 + \sin 2x} d x - \int \frac{\cos x - \sin x}{2 + \sin 2x} d x$ $	ext{આપેલ સમીકરણ સાથે સરખાવતા, } \frac{1}{2 \sqrt{3}} \log |f(x)| = \int \frac{\sin x + \cos x}{2 + \sin 2x} d x$ $	ext{નોંધો કે } 2 + \sin 2x = 3 - (1 - \sin 2x) = 3 - (\sin x - \cos x)^2$ $	ext{ધારો કે } u = \sin x - \cos x, 	ext{ તો } du = (\cos x + \sin x) d x$ $\int \frac{du}{3 - u^2} = \frac{1}{2 \sqrt{3}} \log \left| \frac{\sqrt{3} + u}{\sqrt{3} - u} ight| + C$ $	ext{આમ, } f(x) = \frac{\sqrt{3} + \sin x - \cos x}{\sqrt{3} - \sin x + \cos x}$ $	ext{જ્યારે } x = \frac{\pi}{3}, \sin x = \frac{\sqrt{3}}{2} 	ext{ અને } \cos x = \frac{1}{2}$ $|f(\frac{\pi}{3})| = \left| \frac{\sqrt{3} + \frac{\sqrt{3}}{2} - \frac{1}{2}}{\sqrt{3} - \frac{\sqrt{3}}{2} + \frac{1}{2}} ight| = \frac{\frac{3 \sqrt{3} - 1}{2}}{\frac{\sqrt{3} + 1}{2}} = \frac{3 \sqrt{3} - 1}{\sqrt{3} + 1}$

$\text{જો } \int \frac{1}{\operatorname{cosec} x+\cos x} d x = \frac{1}{2 \sqrt{3}} \log |f(x)| - \int \frac{\cos x-\sin x}{2+\sin 2 x} d x + c, \text{ હોય તો } x = \frac{\pi}{3} \text{ પર } |f(x)| = $

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