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

If $\int \frac{\sin x \cos x}{\sqrt{\cos ^4 x-\sin ^4 x}} d x=-\frac{f(x)}{2}+c$, then domain of $f(x)$ is
  1. $[2 n \pi \cdot(2 n+1) \pi] \cdot n=0,1,2 \ldots$
  2. $\left[(4 n-1) \frac{\pi}{2},(4 n+1) \frac{\pi}{2}\right], n=0,1,2, \ldots$
  3. $\left[(4 n-1) \frac{\pi}{4},(4 n+1) \frac{\pi}{4}\right], n=0,1,2, \ldots$
  4. $\left[\left(2 n \frac{\pi}{4},(2 n+1) \frac{\pi}{4}\right], n=0,1,2, \ldots\right.$

Solution

$I=\int \frac{\sin x \cos x}{\sqrt{\cos ^4 x-\sin ^4 x}} d x \Rightarrow I=\frac{1}{2} \int \frac{\sin 2 x}{\sqrt{\cos 2 x}} d x$ Let $\cos 2 x=t \Rightarrow-2 \sin 2 x d x=d t$ $\therefore I=\frac{-1}{4} \int \frac{1}{\sqrt{t}} d t \Rightarrow I=\frac{-1}{2} \sqrt{t}+C \Rightarrow I=\frac{-\sqrt{\cos 2 x}}{2}+C$
On comparing with $\frac{-f(x)}{2}+C \Rightarrow f(x)=\sqrt{\cos 2 x}$ For domain : $\cos 2 x \geq 0$ $\Rightarrow\left(2 n-\frac{1}{2}\right) \pi \leq 2 x \leq\left(2 n+\frac{1}{2}\right) \pi$ $\Rightarrow(4 n-1) \frac{\pi}{4} \leq x \leq(4 n+1) \frac{\pi}{2}, n=0,1,2, \ldots .$.

Asked in: AP EAMCET 2024 (23 May Shift 1)

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