Question Number 26090 by fahim last updated on 19/Dec/17
$$\int\frac{{a}\mathrm{sin}\:^{\mathrm{3}} \theta+{b}\mathrm{cos}\:^{\mathrm{3}} \theta}{\mathrm{sin}^{\mathrm{2}} \:\theta.\mathrm{cos}^{\mathrm{2}} \:\theta}{d}\theta \\ $$
Answered by Joel578 last updated on 19/Dec/17
$${I}\:=\:\int\:\frac{{a}\mathrm{sin}^{\mathrm{3}} \:\theta}{\mathrm{sin}^{\mathrm{2}} \:\theta\:\mathrm{cos}^{\mathrm{2}} \:\theta}\:{d}\theta\:+\:\int\:\frac{{b}\mathrm{cos}^{\mathrm{3}} \:\theta}{\mathrm{sin}^{\mathrm{2}} \:\theta\:\mathrm{cos}^{\mathrm{2}} \:\theta}\:{d}\theta \\ $$$$\:\:\:\:=\:{a}\:\int\:\:\frac{\mathrm{sin}\:\theta}{\mathrm{cos}^{\mathrm{2}} \:\theta}\:{d}\theta\:+\:{b}\:\int\:\:\frac{\mathrm{cos}\:\theta}{\mathrm{sin}^{\mathrm{2}} \:\theta}\:{d}\theta \\ $$$$\:\:\:\:=\:\frac{{a}}{\mathrm{cos}\:\theta}\:\:−\:\:\frac{{b}}{\mathrm{sin}\:\theta}\:\:+\:\:{C} \\ $$