$\int \frac{\sin ^8 x-\cos ^8 x}{1-2 \sin ^2 x \cos ^2 x}$ (where $\mathrm{C}$ is constant of integration.)

$\int \frac{\sin ^8 x-\cos ^8 x}{1-2 \sin ^2 x \cos ^2 x}$ (where $\mathrm{C}$ is constant of integration.)
  1. $-2 \sin (2 x)+C$
  2. $\frac{1}{2} \cos (2 x)+C$
  3. $2 \cos (2 x)+C$
  4. $-\frac{1}{2} \sin (2 x)+C$

Solution

$\begin{aligned} & \int \frac{\sin ^8 x-\cos ^8 x}{1-2 \sin ^2 x \cos ^2 x} d x \\ & =\int \frac{\left(\sin ^2 x+\cos ^2 x\right)\left(\sin ^2 x-\cos ^2 x\right)\left(1-2 \sin ^2 x \cdot \cos ^2 x\right)}{1-2 \sin ^2 x \cdot \cos ^2 x} d x \\ & =\int\left(\sin ^2 x-\cos ^2 x\right) d x \\ & =\int-\cos 2 x d x=-\frac{1}{2} \sin 2 x+c\end{aligned}$

Asked in: MHT CET 2022 (07 Aug Shift 2)

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