∫((sin^8 x−cos^8 x)/(1−2sin^2 x.cos^2 x)) = ? a) ((−1)/2)sin 2x b)(1/2)sin 2x c)None.


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Question Number 49746 by rahul 19 last updated on 10/Dec/18

$\int \frac{\sin^{8} x - \cos^{8} x}{1 - {2 \sin}^{2} x . \cos^{2} x} = ?$ $a) \frac{- 1}{2} \sin 2 x b) \frac{1}{2} \sin 2 x c) N o n e .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 10/Dec/18
	$sin^8 x−cos^8 x=(sin^4 x+cos^4 x)(sin^4 x−cos^4 x) ={(sin^2 x+cos^2 x)^2 −2sin^2 xcos^2 x}{sin^2 x−cos^2 x} =(1−2sin^2 xcos^2 x)×(−cos2x) ∫((−cos2x)/)dx =((−1)/2)×((sin2x)/)+c$
	${s i n}^{8} x - {c o s}^{8} x = ({s i n}^{4} x + {c o s}^{4} x) ({s i n}^{4} x - {c o s}^{4} x)$ $= {{({s i n}^{2} x + {c o s}^{2} x)}^{2} - 2 {s i n}^{2} {x c o s}^{2} x} {{s i n}^{2} x - {c o s}^{2} x}$ $= (1 - 2 {s i n}^{2} {x c o s}^{2} x) \times (- c o s 2 x)$ $\int \frac{- c o s 2 x}{} d x$ $= \frac{- 1}{2} \times \frac{s i n 2 x}{} + c$
	Commented by rahul 19 last updated on 10/Dec/18
	thank you sir! ��
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