sin^6 x−cos^6 x/sin^2 xcos^2 x


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Question Number 49815 by Aditya789 last updated on 11/Dec/18

$\sin^{6} x - \cos^{6} x / \sin^{2} {xcos}^{2} x$
Commented by Abdo msup. last updated on 12/Dec/18

$i f y o u m e a n s i m i f i c a t i o n$ $A = \frac{{({s i n}^{2} x)}^{3} - {({c o s}^{2} x)}^{3}}{{s i n}^{2} x . {c o s}^{2} x} = \frac{({s i n}^{2} x - {c o s}^{2} x) ({s i n}^{4} x + {s i n}^{2} {x c o s}^{2} x + {c o s}^{4} x)}{{s i n}^{2} x . {c o s}^{2} x}$ $= \frac{- 2 c o s (2 x) ({({s i n}^{2} x + {c o s}^{2} x)}^{2} - {s i n}^{2} x {c o s}^{2} x)}{{s i n}^{2} x {c o s}^{2} x}$ $= - 2 c o s (2 x) (\frac{4}{{s i n}^{2} (2 x)} - 1)$ $= \frac{8 c o s (2 x)}{{s i n}^{2} (2 x)} + 2 c o s (2 x) .$
Commented by Abdo msup. last updated on 12/Dec/18

$A = - \frac{8 c o s (2 x)}{{s i n}^{2} (2 x)} + 2 c o s (2 x) .$
Commented by Abdo msup. last updated on 15/Dec/18

$\int A (x) d x = - 4 \int \frac{2 c o s (2 x)}{{s i n}^{2} (2 x)} d x + 2 \int c o s (2 x) d x$ $= \frac{4}{s i n (2 x)} + s i n (2 x) + c$
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