int frac{1+sqrt{3} cot x}{1-sqrt{3} cot x} d | vedclass

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Question

(A) $\int \frac{1+\sqrt{3} \cot x}{1-\sqrt{3} \cot x} d x = \int \frac{	an x+\sqrt{3}}{	an x-\sqrt{3}} d x = \int \frac{\sin x+\sqrt{3} \cos x}{\sin

Vedclass · Accepted Answer

$-\frac{x}{2}+\frac{\sqrt{3}}{2} \log \left|\sin \left(x-\frac{\pi}{3}\right)\right|+c$

Vedclass · Answer

(A) $\int \frac{1+\sqrt{3} \cot x}{1-\sqrt{3} \cot x} d x = \int \frac{	an x+\sqrt{3}}{	an x-\sqrt{3}} d x = \int \frac{\sin x+\sqrt{3} \cos x}{\sin x-\sqrt{3} \cos x} d x$माना $\sin x+\sqrt{3} \cos x = K_1(\cos x+\sqrt{3} \sin x) + K_2(\sin x-\sqrt{3} \cos x)$.$\sin x$ और $\cos x$ के गुणांकों की तुलना करने पर:$\sqrt{3} K_1 + K_2 = 1$ $(i)$$K_1 - \sqrt{3} K_2 = \sqrt{3}$ (ii)इन्हें हल करने पर, हमें $K_1 = \frac{\sqrt{3}}{2}$ और $K_2 = -\frac{1}{2}$ प्राप्त होता है।अतः, समाकलन $\int \frac{K_1(\cos x+\sqrt{3} \sin x) + K_2(\sin x-\sqrt{3} \cos x)}{\sin x-\sqrt{3} \cos x} d x$ बन जाता है।$= K_1 \int \frac{d(\sin x-\sqrt{3} \cos x)}{\sin x-\sqrt{3} \cos x} + K_2 \int 1 d x$$= \frac{\sqrt{3}}{2} \ln |\sin x-\sqrt{3} \cos x| - \frac{1}{2} x + C$चूंकि $\sin x-\sqrt{3} \cos x = 2(\frac{1}{2} \sin x - \frac{\sqrt{3}}{2} \cos x) = 2 \sin(x-\frac{\pi}{3})$,अतः व्यंजक $-\frac{x}{2} + \frac{\sqrt{3}}{2} \ln |2 \sin(x-\frac{\pi}{3})| + C = -\frac{x}{2} + \frac{\sqrt{3}}{2} \ln |\sin(x-\frac{\pi}{3})| + C'$ हो जाता है।

$\int \frac{1+\sqrt{3} \cot x}{1-\sqrt{3} \cot x} d x=$

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