If int frac{2 , dx}{sqrt{cot^2 x - tan^2 x}} | vedclass

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(C) We have,$\int \frac{2 \, dx}{\sqrt{\cot^2 x - 	an^2 x}}$.Converting to $\sin x$ and $\cos x$:$= \int \frac{2 \, dx}{\sqrt{\frac{\cos^2 x}{\sin^2

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$\cos 2x$

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(C) We have,$\int \frac{2 \, dx}{\sqrt{\cot^2 x - 	an^2 x}}$.Converting to $\sin x$ and $\cos x$:$= \int \frac{2 \, dx}{\sqrt{\frac{\cos^2 x}{\sin^2 x} - \frac{\sin^2 x}{\cos^2 x}}} = \int \frac{2 \sin x \cos x}{\sqrt{\cos^4 x - \sin^4 x}} \, dx$.Using $\sin 2x = 2 \sin x \cos x$ and $\cos^4 x - \sin^4 x = (\cos^2 x - \sin^2 x)(\cos^2 x + \sin^2 x) = \cos 2x \cdot 1 = \cos 2x$:$= \int \frac{\sin 2x \, dx}{\sqrt{\cos 2x}}$.Let $t = \cos 2x$,then $dt = -2 \sin 2x \, dx$,which implies $\sin 2x \, dx = -\frac{1}{2} dt$.$= -\frac{1}{2} \int \frac{dt}{\sqrt{t}} = -\frac{1}{2} \int t^{-1/2} \, dt = -\frac{1}{2} \left( \frac{t^{1/2}}{1/2} ight) + c = -\sqrt{t} + c$.Substituting back $t = \cos 2x$,we get $-\sqrt{\cos 2x} + c$.Comparing with $-\sqrt{f(x)} + c$,we find $f(x) = \cos 2x$.

If $\int \frac{2 \, dx}{\sqrt{\cot^2 x - \tan^2 x}} = -\sqrt{f(x)} + c$,then $f(x) =$

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