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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

{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! ��