Question Number 18590 by Joel577 last updated on 25/Jul/17
$$\int\:\frac{\mathrm{sin}\:{x}}{\mathrm{1}\:+\:\mathrm{cos}^{\mathrm{2}} \:{x}}\:{dx} \\ $$
Answered by Arnab Maiti last updated on 25/Jul/17
$$\mathrm{put}\:\mathrm{cos}\:\mathrm{x}=\mathrm{z} \\ $$$$\:\:\:\:\:−\mathrm{sin}\:\mathrm{x}\:\mathrm{dx}=\mathrm{dz} \\ $$$$=\int\frac{−\mathrm{dz}}{\mathrm{1}+\mathrm{z}^{\mathrm{2}} } \\ $$$$=−\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{z}\right)+\mathrm{C} \\ $$$$=−\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{cos}\:\mathrm{x}\right)+\mathrm{C} \\ $$
Commented by Joel577 last updated on 25/Jul/17
$${thank}\:{you}\:{very}\:{much} \\ $$