If ∫sin8x−cos8x1−2sin2x cos2x dx=f(x)+c, f(π4)=−12, and Ltx→0 2(x+f(x))xn is non–zero finite and equal to l then n−l=

sin8x−cos8x1−2 sin2x cos2x=(sin4x−cos4x)(sin4x+cos4x)sin4x+cos4x=sin4x−cos4x=−cos2x

If ∫sin8x−cos8x1−2sin2x cos2x dx=f(x)+c, f(π4)=−12, and Ltx→0 2(x+f(x))xn is non–zero finite and equal to l then n−l=

sin8x−cos8x1−2 sin2x cos2x=(sin4x−cos4x)(sin4x+cos4x)sin4x+cos4x=sin4x−cos4x=−cos2x

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If $\int \frac{\sin^{8} x - \cos^{8} x}{1 - 2 \sin^{2} x \cos^{2} x} d x = f (x) + c, f (\frac{π}{4}) = \frac{- 1}{2},$ and $\underset{x \to 0}{L t} \frac{2 (x + f (x))}{x^{n}}$ is non–zero finite and equal to $l$ then $n - l =$

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$\frac{5}{3}$

$\frac{- 7}{2}$

answer is A.

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$\begin{array}{l} \frac{\sin^{8} x - \cos^{8} x}{1 - 2 \sin^{2} x \cos^{2} x} \\ = \frac{(\sin^{4} x - \cos^{4} x) (\sin^{4} x + \cos^{4} x)}{\sin^{4} x + \cos^{4} x} \\ = \sin^{4} x - \cos^{4} x = - \cos 2 x \end{array}$

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If ∫sin8x−cos8x1−2sin2x cos2x dx=f(x)+c, f(π4)=−12, and Ltx→0 2(x+f(x))xn is non–zero finite and equal to l then n−l=

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If $\int \frac{\sin^{8} x - \cos^{8} x}{1 - 2 \sin^{2} x \cos^{2} x} d x = f (x) + c, f (\frac{π}{4}) = \frac{- 1}{2},$ and $\underset{x \to 0}{L t} \frac{2 (x + f (x))}{x^{n}}$ is non–zero finite and equal to $l$ then $n - l =$