If int frac{sin x cos x}{sqrt{cos^4 x - sin^4 | vedclass

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(C) Given integral $I = \int \frac{\sin x \cos x}{\sqrt{\cos^4 x - \sin^4 x}} dx$.Using $\sin 2x = 2 \sin x \cos x$ and $\cos^4 x - \sin^4 x = (\cos^2

Vedclass · Accepted Answer

$[(4n-1)\frac{\pi}{4}, (4n+1)\frac{\pi}{4}], n=0, 1, 2, \ldots$

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(C) Given integral $I = \int \frac{\sin x \cos x}{\sqrt{\cos^4 x - \sin^4 x}} dx$.Using $\sin 2x = 2 \sin x \cos x$ and $\cos^4 x - \sin^4 x = (\cos^2 x - \sin^2 x)(\cos^2 x + \sin^2 x) = \cos 2x$.So,$I = \int \frac{\frac{1}{2} \sin 2x}{\sqrt{\cos 2x}} dx = \frac{1}{2} \int \sin 2x (\cos 2x)^{-1/2} dx$.Let $t = \cos 2x$,then $dt = -2 \sin 2x dx$,so $\sin 2x dx = -\frac{1}{2} dt$.$I = \frac{1}{2} \int -\frac{1}{2} t^{-1/2} dt = -\frac{1}{4} \cdot 2 t^{1/2} + c = -\frac{1}{2} \sqrt{\cos 2x} + c$.Comparing with $-\frac{f(x)}{2} + c$,we get $f(x) = \sqrt{\cos 2x}$.For the domain,$\cos 2x \geq 0$.This implies $2n\pi - \frac{\pi}{2} \leq 2x \leq 2n\pi + \frac{\pi}{2}$.Dividing by $2$,we get $n\pi - \frac{\pi}{4} \leq x \leq n\pi + \frac{\pi}{4}$.For $n=0$,$[-\frac{\pi}{4}, \frac{\pi}{4}]$. For $n=1$,$[\frac{3\pi}{4}, \frac{5\pi}{4}]$.This matches the general form $[(4n-1)\frac{\pi}{4}, (4n+1)\frac{\pi}{4}]$.

If $\int \frac{\sin x \cos x}{\sqrt{\cos^4 x - \sin^4 x}} dx = -\frac{f(x)}{2} + c$,then the domain of $f(x)$ is

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