If int frac{d x}{cos ^4 x+sin ^4 x}=frac{1}{s | vedclass

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(A) Given $I = \int \frac{d x}{\cos ^4 x+\sin ^4 x}$.Dividing numerator and denominator by $\cos^4 x$,we get $I = \int \frac{\sec^4 x}{1+	an^4 x} d x

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$\frac{	an x-\cot x}{\sqrt{2}}$

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(A) Given $I = \int \frac{d x}{\cos ^4 x+\sin ^4 x}$.Dividing numerator and denominator by $\cos^4 x$,we get $I = \int \frac{\sec^4 x}{1+	an^4 x} d x = \int \frac{\sec^2 x(1+	an^2 x)}{1+	an^4 x} d x$.Let $	an x = t$,then $\sec^2 x d x = d t$.$I = \int \frac{1+t^2}{1+t^4} d t = \int \frac{1+\frac{1}{t^2}}{t^2+\frac{1}{t^2}} d t$.$I = \int \frac{1+\frac{1}{t^2}}{(t-\frac{1}{t})^2+2} d t$.Let $u = t-\frac{1}{t}$,then $d u = (1+\frac{1}{t^2}) d t$.$I = \int \frac{d u}{u^2+(\sqrt{2})^2} = \frac{1}{\sqrt{2}} 	an^{-1}(\frac{u}{\sqrt{2}}) + C$.Substituting $u = t-\frac{1}{t} = 	an x - \cot x$,we get $I = \frac{1}{\sqrt{2}} 	an^{-1}(\frac{	an x - \cot x}{\sqrt{2}}) + C$.Comparing this with the given form,$g(x) = \frac{	an x - \cot x}{\sqrt{2}}$.

If $\int \frac{d x}{\cos ^4 x+\sin ^4 x}=\frac{1}{\sqrt{2}} \tan ^{-1}[g(x)]+C$,then $g(x)$ equals

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