int frac{d x}{sin left(x-frac{pi}{3}right) co | vedclass

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Question

(A) माना $I = \int \frac{dx}{\sin(x - \frac{\pi}{3}) \cos x}$ है।सूत्र $\sin(A - B) = \sin A \cos B - \cos A \sin B$ का उपयोग करने पर,हमें $\sin(x - \

Vedclass · Accepted Answer

$2 \log \left| 	an x - \sqrt{3} ight| + C$

Vedclass · Answer

(A) माना $I = \int \frac{dx}{\sin(x - \frac{\pi}{3}) \cos x}$ है।सूत्र $\sin(A - B) = \sin A \cos B - \cos A \sin B$ का उपयोग करने पर,हमें $\sin(x - \frac{\pi}{3}) = \sin x \cos(\frac{\pi}{3}) - \cos x \sin(\frac{\pi}{3}) = \frac{1}{2} \sin x - \frac{\sqrt{3}}{2} \cos x$ प्राप्त होता है।इस मान को समाकलन में प्रतिस्थापित करने पर:$I = \int \frac{dx}{(\frac{1}{2} \sin x - \frac{\sqrt{3}}{2} \cos x) \cos x} = \int \frac{dx}{\cos^2 x (\frac{1}{2} 	an x - \frac{\sqrt{3}}{2})} = \int \frac{\sec^2 x dx}{\frac{1}{2} (	an x - \sqrt{3})} = 2 \int \frac{\sec^2 x dx}{	an x - \sqrt{3}}$।माना $u = 	an x - \sqrt{3}$,तब $du = \sec^2 x dx$ है।$I = 2 \int \frac{du}{u} = 2 \log |u| + C = 2 \log |	an x - \sqrt{3}| + C$।

$\int \frac{d x}{\sin \left(x-\frac{\pi}{3}\right) \cos x} = $

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