यदि int frac{dx}{2 cos x + 3 sin x + 4} = fra | vedclass

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(C) माना $I = \int \frac{dx}{2 \cos x + 3 \sin x + 4}$.Weierstrass प्रतिस्थापन का उपयोग करते हुए,$t = 	an\left(\frac{x}{2}ight)$ लें,तो $dx = \frac

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$\frac{5 \pi}{12}$

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(C) माना $I = \int \frac{dx}{2 \cos x + 3 \sin x + 4}$.Weierstrass प्रतिस्थापन का उपयोग करते हुए,$t = 	an\left(\frac{x}{2}ight)$ लें,तो $dx = \frac{2 dt}{1+t^2}$,$\cos x = \frac{1-t^2}{1+t^2}$,और $\sin x = \frac{2t}{1+t^2}$.इन मानों को समाकलन में रखने पर:$I = \int \frac{\frac{2 dt}{1+t^2}}{2\left(\frac{1-t^2}{1+t^2}ight) + 3\left(\frac{2t}{1+t^2}ight) + 4} = \int \frac{2 dt}{2 - 2t^2 + 6t + 4 + 4t^2} = \int \frac{2 dt}{2t^2 + 6t + 6} = \int \frac{dt}{t^2 + 3t + 3}$.हर में पूर्ण वर्ग बनाने पर: $t^2 + 3t + 3 = \left(t + \frac{3}{2}ight)^2 + \frac{3}{4}$.अतः,$I = \int \frac{dt}{\left(t + \frac{3}{2}ight)^2 + \left(\frac{\sqrt{3}}{2}ight)^2} = \frac{1}{\frac{\sqrt{3}}{2}} 	an^{-1}\left(\frac{t + 3/2}{\sqrt{3}/2}ight) + c = \frac{2}{\sqrt{3}} 	an^{-1}\left(\frac{2t+3}{\sqrt{3}}ight) + c$.इसे दिए गए रूप $\frac{2}{\sqrt{3}} f(x) + c$ के साथ तुलना करने पर,हमें $f(x) = 	an^{-1}\left(\frac{2 	an(x/2) + 3}{\sqrt{3}}ight)$ प्राप्त होता है।अब,$f\left(\frac{2 \pi}{3}ight) = 	an^{-1}\left(\frac{2 	an(\pi/3) + 3}{\sqrt{3}}ight) = 	an^{-1}\left(\frac{2\sqrt{3} + 3}{\sqrt{3}}ight) = 	an^{-1}(2 + \sqrt{3})$.चूंकि $	an(75^\circ) = 	an\left(\frac{5 \pi}{12}ight) = 2 + \sqrt{3}$,इसलिए $f\left(\frac{2 \pi}{3}ight) = \frac{5 \pi}{12}$.

यदि $\int \frac{dx}{2 \cos x + 3 \sin x + 4} = \frac{2}{\sqrt{3}} f(x) + c$ है,तो $f\left(\frac{2 \pi}{3}\right) =$

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