Question Number 95009 by bobhans last updated on 22/May/20
$$\int\:\:\frac{{e}^{{x}} +\mathrm{2}{xe}^{{x}} \mathrm{sin}\:{x}−\mathrm{2}{xe}^{{x}} \mathrm{cos}\:{x}}{\:\sqrt{{x}}\:\mathrm{sin}\:^{\mathrm{2}} {x}}\:{dx}\: \\ $$
Commented by bobhans last updated on 22/May/20
my children exam
Commented by M±th+et+s last updated on 22/May/20
$${think}\:{that}\:{its} \\ $$$$\int\frac{{e}^{{x}} {sin}\left({x}\right)+\mathrm{2}{xe}^{{x}} {sin}\left({x}\right)−\mathrm{2}{xe}^{{x}} {cos}\left({x}\right)}{\:\sqrt{{x}}{sin}^{\mathrm{2}} \left({x}\right)}{dx} \\ $$
Commented by bobhans last updated on 23/May/20
$$\mathrm{yes}.\:\mathrm{if}\:\int\:\frac{\mathrm{e}^{\mathrm{x}} \mathrm{sin}\:\left(\mathrm{x}\right)−\mathrm{2xe}^{\mathrm{x}} \mathrm{sin}\:\left(\mathrm{x}\right)−\mathrm{2xe}^{\mathrm{x}} \mathrm{cos}\:\left(\mathrm{x}\right)}{\:\sqrt{\mathrm{x}}\:\mathrm{sin}\:^{\mathrm{2}} \left(\mathrm{x}\right)}\:\mathrm{dx}\: \\ $$$$\mathrm{can}\:\mathrm{be}\:\mathrm{solved}\:.\: \\ $$$$\int\:\frac{\mathrm{d}}{\mathrm{dx}}\:\left[\frac{\mathrm{2}\sqrt{\mathrm{x}}\:\mathrm{e}^{\mathrm{x}} }{\mathrm{sin}\:\left(\mathrm{x}\right)}\:\right]\:=\:\frac{\mathrm{2}\sqrt{\mathrm{x}}\:\mathrm{e}^{\mathrm{x}} }{\mathrm{sin}\:\left(\mathrm{x}\right)}\:+\:\mathrm{c}\: \\ $$