Question Number 49761 by rahul 19 last updated on 10/Dec/18
$${Calculate}\:: \\ $$$$\:\int\frac{\:\:\mathrm{sin}^{\mathrm{2}} {x}\:\mathrm{cos}^{\mathrm{2}} {x}}{\left(\mathrm{sin}^{\mathrm{3}} {x}+\mathrm{cos}^{\mathrm{3}} {x}\right)^{\mathrm{2}} }\:{dx} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 10/Dec/18
$$\int\frac{{tan}^{\mathrm{2}} {xcos}^{\mathrm{4}} {x}}{{cos}^{\mathrm{6}} {x}\left(\mathrm{1}+{tan}^{\mathrm{3}} {x}\right)^{\mathrm{3}} }{dx} \\ $$$$\int\frac{{tan}^{\mathrm{2}} {xsec}^{\mathrm{2}} {x}}{\left(\mathrm{1}+{tan}^{\mathrm{3}} {x}\right)^{\mathrm{3}} }{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\int\frac{{d}\left(\mathrm{1}+{tan}^{\mathrm{3}} {x}\right)}{\left(\mathrm{1}+{tan}^{\mathrm{3}} {x}\right)^{\mathrm{3}} }{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}×\frac{\left(\mathrm{1}+{tan}^{\mathrm{3}} {x}\right)^{−\mathrm{2}} }{−\mathrm{2}}=\frac{\mathrm{1}}{−\mathrm{6}}×\frac{\mathrm{1}}{\left(\mathrm{1}+{tan}^{\mathrm{3}} {x}\right)^{\mathrm{2}} }+{c} \\ $$
Commented by rahul 19 last updated on 10/Dec/18
thank you sir!��