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

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(B) Let $I = \int \frac{dx}{\cos^3 x (	an^3 x + 1)} = \int \frac{\sec^3 x}{	an^3 x + 1} dx$. Substitute $u = 	an x$,then $du = \sec^2 x dx$. $I = \

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

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(B) Let $I = \int \frac{dx}{\cos^3 x (	an^3 x + 1)} = \int \frac{\sec^3 x}{	an^3 x + 1} dx$. Substitute $u = 	an x$,then $du = \sec^2 x dx$. $I = \int \frac{\sec x}{	an^3 x + 1} du = \int \frac{\sqrt{1+u^2}}{u^3+1} du$. Alternatively,divide numerator and denominator by $\cos^3 x$: $I = \int \frac{\sec^2 x \sec x}{	an^3 x + 1} dx$. Using the substitution $t = \sin x - \cos x$,we have $dt = (\cos x + \sin x) dx$. Also $t^2 = 1 - 2\sin x \cos x$,so $\sin x \cos x = \frac{1-t^2}{2}$. $\sin^3 x + \cos^3 x = (\sin x + \cos x)(1 - \sin x \cos x) = (\sin x + \cos x)(1 - \frac{1-t^2}{2}) = (\sin x + \cos x)(\frac{1+t^2}{2})$. Since $dt = (\sin x + \cos x) dx$,we get $dx = \frac{dt}{\sin x + \cos x}$. $I = \int \frac{dt}{(\sin x + \cos x)^2 (\frac{1+t^2}{2})} = \int \frac{2 dt}{(1+2\sin x \cos x)(1+t^2)} = \int \frac{2 dt}{(1 + 1 - t^2)(1+t^2)} = \int \frac{2 dt}{(2-t^2)(1+t^2)}$. Using partial fractions: $\frac{2}{(2-t^2)(1+t^2)} = \frac{A'}{2-t^2} + \frac{B'}{1+t^2}$. $2 = A'(1+t^2) + B'(2-t^2) \implies A' = \frac{2}{3}, B' = \frac{2}{3}$. $I = \frac{2}{3} \int \frac{dt}{2-t^2} + \frac{2}{3} \int \frac{dt}{1+t^2} = \frac{2}{3} \cdot \frac{1}{2\sqrt{2}} \log \left|\frac{\sqrt{2}+t}{\sqrt{2}-t}ight| + \frac{2}{3} 	an^{-1}(t) + c$. $I = \frac{1}{3\sqrt{2}} \log \left|\frac{\sqrt{2}+t}{\sqrt{2}-t}ight| + \frac{2}{3} 	an^{-1}(t) + c$. Comparing with $A \log \left|\frac{\sqrt{2}+t}{\sqrt{2}-t}ight| + B 	an^{-1}(t) + c$,we get $A = \frac{1}{3\sqrt{2}}$ and $B = \frac{2}{3}$. Thus,$\frac{B}{A} = \frac{2/3}{1/(3\sqrt{2})} = 2\sqrt{2}$. Therefore,$(\frac{B}{A}, t) = (2\sqrt{2}, \sin x - \cos x)$.

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

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