यदि int frac{1}{cos^3 x sqrt{sin 2x}} dx = p( | vedclass

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(C) दिया गया समाकल $I = \int \frac{1}{\cos^3 x \sqrt{2 \sin x \cos x}} dx = \frac{1}{\sqrt{2}} \int \frac{1}{\cos^4 x \sqrt{	an x}} dx$$I = \frac{1}{

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$p = \frac{\sqrt{2}}{5}, q = 5$

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(C) दिया गया समाकल $I = \int \frac{1}{\cos^3 x \sqrt{2 \sin x \cos x}} dx = \frac{1}{\sqrt{2}} \int \frac{1}{\cos^4 x \sqrt{	an x}} dx$$I = \frac{1}{\sqrt{2}} \int \frac{\sec^4 x}{\sqrt{	an x}} dx = \frac{1}{\sqrt{2}} \int \frac{(1+	an^2 x) \sec^2 x}{\sqrt{	an x}} dx$माना $	an x = t$, तब $\sec^2 x dx = dt$$I = \frac{1}{\sqrt{2}} \int (1+t^2) t^{-1/2} dt = \frac{1}{\sqrt{2}} \int (t^{-1/2} + t^{3/2}) dt$$I = \frac{1}{\sqrt{2}} \left[ 2t^{1/2} + \frac{2}{5} t^{5/2} ight] + c = \sqrt{2} \sqrt{	an x} + \frac{\sqrt{2}}{5} 	an^2 x \sqrt{	an x} + c = \frac{\sqrt{2}}{5} (	an^2 x + 5) \sqrt{	an x} + c$$p(	an^2 x + q)\sqrt{	an x} + c$ से तुलना करने पर, $p = \frac{\sqrt{2}}{5}$ और $q = 5$ प्राप्त होता है।

यदि $\int \frac{1}{\cos^3 x \sqrt{\sin 2x}} dx = p(\tan^2 x + q)\sqrt{\tan x} + c$ है, तो $p$ और $q$ के मान क्रमशः हैं

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