Question Number 211155 by Spillover last updated on 30/Aug/24
$$\int\frac{\left(\mathrm{sin}\:^{{n}} \left(\theta\right)−\mathrm{sin}\:\left(\theta\right)\right)^{\frac{\mathrm{1}}{{n}}} \mathrm{cos}\:\left(\theta\right)}{\mathrm{sin}\:^{{n}+\mathrm{1}} \left(\theta\right)}{d}\theta \\ $$$$ \\ $$
Answered by Frix last updated on 30/Aug/24
$$=\int\frac{{d}\theta}{\mathrm{sin}\:\theta}\:−\int\left(\mathrm{sin}\:\theta\right)^{\frac{\mathrm{1}−{n}−{n}^{\mathrm{2}} }{{n}}} \mathrm{cos}\:\theta\:{d}\theta= \\ $$$$=\mathrm{ln}\:\mid\frac{\mathrm{1}−\mathrm{cos}\:\theta}{\mathrm{sin}\:\theta}\mid\:+\frac{{n}}{{n}^{\mathrm{2}} −\mathrm{1}}\left(\mathrm{sin}\:\theta\right)^{\frac{\mathrm{1}−{n}^{\mathrm{2}} }{{n}}} +{C} \\ $$