Question Number 49746 by rahul 19 last updated on 10/Dec/18
$$\int\frac{\mathrm{sin}^{\mathrm{8}} {x}−\mathrm{cos}^{\mathrm{8}} {x}}{\mathrm{1}−\mathrm{2sin}^{\mathrm{2}} {x}.\mathrm{cos}^{\mathrm{2}} {x}}\:=\:? \\ $$$$\left.{a}\left.\right)\left.\:\frac{−\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\mathrm{2}{x}\:\:\:{b}\right)\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\mathrm{2}{x}\:\:\:{c}\right){None}. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 10/Dec/18
$${sin}^{\mathrm{8}} {x}−{cos}^{\mathrm{8}} {x}=\left({sin}^{\mathrm{4}} {x}+{cos}^{\mathrm{4}} {x}\right)\left({sin}^{\mathrm{4}} {x}−{cos}^{\mathrm{4}} {x}\right) \\ $$$$=\left\{\left({sin}^{\mathrm{2}} {x}+{cos}^{\mathrm{2}} {x}\right)^{\mathrm{2}} −\mathrm{2}{sin}^{\mathrm{2}} {xcos}^{\mathrm{2}} {x}\right\}\left\{{sin}^{\mathrm{2}} {x}−{cos}^{\mathrm{2}} {x}\right\} \\ $$$$=\left(\mathrm{1}−\mathrm{2}{sin}^{\mathrm{2}} {xcos}^{\mathrm{2}} {x}\right)×\left(−{cos}\mathrm{2}{x}\right) \\ $$$$\int\frac{−{cos}\mathrm{2}{x}}{}{dx} \\ $$$$=\frac{−\mathrm{1}}{\mathrm{2}}×\frac{{sin}\mathrm{2}{x}}{}+{c} \\ $$$$ \\ $$
Commented by rahul 19 last updated on 10/Dec/18
thank you sir! ��