Mathematics

$\displaystyle \int { \frac { \sin ^{ 8 }{ x } -\cos ^{ 8 }{ x } }{ 1-2\sin ^{ 2 }{ x } \cos ^{ 2 }{ x } } } dx$

$\dfrac {1}{2}\sin 2x+C$

SOLUTION
$\int{\cfrac{\sin^{8}{x} - \cos^{8}{x}}{1 - 2 \sin^{2}{\cos^{2}{x}}} dx}$
$= \int{\cfrac{\left( \sin^{4}{x} - \cos^{4}{x} \right) \left( \sin^{4}{x} + \cos^{4}{x} \right)}{1 - 2 \sin^{2}{x} \cos^{2}{x}} dx}$
$= \int{\left( \cfrac{\left[ \left( \sin^{2}{x} - \cos^{2}{x} \right) \left( \sin^{2}{x} + \cos^{2}{x} \right) \right] \left[ {\left( \sin^{2}{x} + \cos^{2}{x} \right)}^{2} - 2 \sin^{2}{x} \cos^{2}{x} \right]}{1 - 2 \sin^{2}{x} \cos^{2}{x}} \right) dx}$
$= \int{ \left( \cfrac{\left( \cos{2x} \right) \left( 1 - 2 \sin^{2}{x} \cos^{2}{x} \right)}{\left( 1 - 2 \sin^{2}{x} \cos^{2}{x} \right)} \right) dx} \quad \left( \because \cos^{2}{x} - \sin^{2}{x} = \cos{2x}; \text{ and } \sin^{2}{x} + \cos^{2}{x} = 1 \right)$
$= \int{\cos{2x} \; dx}$
$= \cfrac{\sin{2x}}{2} + C$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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