MCQ
$\int {\frac{{{{\sin }^8}\,x - {{\cos }^8}\,x}}{{\left( {1 - 2\,{{\sin }^2}\,x\,{{\cos }^2}\,x} \right)}}} dx$ is equal to
- A$\frac{1}{2}\,\sin \,2x + c$
- ✓$-\frac{1}{2}\,\sin \,2x + c$
- C$-\frac{1}{2}\,\sin \,x + c$
- D$ - \sin {^2}\,x + c$
$ = \int {\frac{{{{\left( {{{\sin }^4}x} \right)}^2} - {{\left( {{{\cos }^4}x} \right)}^2}}}{{1 - 2{{\sin }^2}x{{\cos }^2}x}}} .dx$
$ = \int {\frac{{\left( {{{\sin }^4}x + {{\cos }^4}x} \right)\left( {{{\sin }^4}x - {{\cos }^4}x} \right)}}{{1 - 2{{\sin }^2}x{{\cos }^2}x}}} .dx$
$\left[ {{{\left( {{{\sin }^2}x + {{\cos }^2}x} \right)}^2} - 2{{\sin }^2}x{{\cos }^2}x} \right]$
$ = \int {\frac{{\left[ {\left( {{{\sin }^2}x + {{\cos }^2}x} \right]\left[ {{{\sin }^2}x - {{\cos }^2}x} \right]} \right.}}{{1 - 2{{\sin }^2}x{{\cos }^2}x}}} dx$
$ = - \int {\cos } 2xdx = \frac{{ - \sin 2x}}{2} + c$
$ = - \frac{1}{2}\sin 2x + c$
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$\sqrt{1-\pi^2}-1$
$\frac{\pi}{2}-1$
$\frac{\pi}{2}+1$
${\pi}+{1}$
$($ where $det(B)$ denotes determinant of Matrix $B) -$