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Question: Answered & Verified by Expert
$\int \frac{\sin ^8 x-\cos ^8 x}{1-2 \sin ^2 x \cos ^2 x} d x=$
MathematicsIndefinite IntegrationTS EAMCETTS EAMCET 2023 (12 May Shift 1)
Options:
  • A $\frac{1}{2} \cos 2 x+c$
  • B $\frac{-1}{2} \cos 2 x+c$
  • C $\frac{-1}{(1+\tan x)^2}+c$
  • D $\frac{-1}{2} \sin 2 x+c$
Solution:
1238 Upvotes Verified Answer
The correct answer is: $\frac{-1}{2} \sin 2 x+c$
$\int \frac{\sin ^8 x-\cos ^8 x}{1-2 \sin ^2 x \cos ^2 x} d x$
$\begin{aligned} & =\int \frac{\left(\sin ^4 x+\cos ^4 x\right)\left(\sin ^2 x+\cos ^2 x\right)\left(\sin ^2 x-\cos ^2 x\right)}{\left(\sin ^4 x+\cos ^4 x\right)} d x \\ & =\int\left(\sin ^2 x-\cos ^2 x\right) d x=-\int \cos 2 x d x \\ & =-\frac{1}{2} \sin 2 x+C .\end{aligned}$

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