यदि int frac{dx}{sin^3 x + cos^3 x} = A log l | vedclass

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(B) माना $I = \int \frac{dx}{\cos^3 x (	an^3 x + 1)} = \int \frac{\sec^3 x}{	an^3 x + 1} dx$. $u = 	an x$ प्रतिस्थापन करने पर,$du = \sec^2 x dx$. अ

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$(2\sqrt{2}, \sin x - \cos x)$

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(B) माना $I = \int \frac{dx}{\cos^3 x (	an^3 x + 1)} = \int \frac{\sec^3 x}{	an^3 x + 1} dx$. $u = 	an x$ प्रतिस्थापन करने पर,$du = \sec^2 x dx$. अंश और हर को $\cos^3 x$ से विभाजित करने पर: $t = \sin x - \cos x$ प्रतिस्थापन करने पर,$dt = (\cos x + \sin x) dx$. $t^2 = 1 - 2\sin x \cos x$,अतः $\sin x \cos x = \frac{1-t^2}{2}$. $\sin^3 x + \cos^3 x = (\sin x + \cos x)(1 - \sin x \cos x) = (\sin x + \cos x)(\frac{1+t^2}{2})$. $I = \int \frac{2 dt}{(2-t^2)(1+t^2)}$. आंशिक भिन्न का उपयोग करने पर: $\frac{2}{(2-t^2)(1+t^2)} = \frac{2/3}{2-t^2} + \frac{2/3}{1+t^2}$. $I = \frac{2}{3} \int \frac{dt}{2-t^2} + \frac{2}{3} \int \frac{dt}{1+t^2} = \frac{1}{3\sqrt{2}} \log \left|\frac{\sqrt{2}+t}{\sqrt{2}-t}ight| + \frac{2}{3} 	an^{-1}(t) + c$. तुलना करने पर $A = \frac{1}{3\sqrt{2}}$ और $B = \frac{2}{3}$. अतः $\frac{B}{A} = 2\sqrt{2}$. इस प्रकार,$(\frac{B}{A}, t) = (2\sqrt{2}, \sin x - \cos x)$.

यदि $\int \frac{dx}{\sin^3 x + \cos^3 x} = A \log \left|\frac{\sqrt{2}+t}{\sqrt{2}-t}\right| + B \tan^{-1}(t) + c$ है,तो $\left(\frac{B}{A}, t\right) =$

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