_ ^ ^8 x - ^8 x 1 - 2 ^2 x ^2 x ;dx = - Vidyadip

MCQ

$\int_{}^{} {\frac{{{{\sin }^8}x - {{\cos }^8}x}}{{1 - 2{{\sin }^2}x{{\cos }^2}x}}\;dx = } $

A
$\sin 2x + c$
✓
$ - \frac{1}{2}\sin 2x + c$
C
$\frac{1}{2}\sin 2x + c$
D
$ - \sin 2x + c$

✓

Answer

Correct option: B.

$ - \frac{1}{2}\sin 2x + c$

b
(b)$\int_{}^{} {\frac{{{{\sin }^8}x - {{\cos }^8}x}}{{1 - 2{{\sin }^2}x{{\cos }^2}x}}\,dx} $
$ = \int_{}^{} {\frac{{({{\sin }^4}x + {{\cos }^4}x)({{\sin }^4}x - {{\cos }^4}x)}}{{{{({{\sin }^2}x + {{\cos }^2}x)}^2} - 2{{\sin }^2}x{{\cos }^2}x}}} \,dx$
$ = \int_{}^{} {({{\sin }^4}x - {{\cos }^4}x)\,dx} $
$ = \int_{}^{} {({{\sin }^2}x + {{\cos }^2}x)({{\sin }^2}x - {{\cos }^2}x)\,dx} $
$ = \int_{}^{} {({{\sin }^2}x - {{\cos }^2}x)\,dx} $$ = \int_{}^{} { - \cos 2x\,dx = - \frac{{\sin 2x}}{2} + c.} $

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