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Question Number 196276 by SaRahAli last updated on 21/Aug/23

Answered by MM42 last updated on 21/Aug/23

∫ (dx/(1+sin2x)) =∫ ((1+tan^2 x)/((1+tanx)^2 )) dx  =^(1+tanx=u)  ∫ (du/u) =lnu+c  =ln(1+tanx)+c ✓

$$\int\:\frac{{dx}}{\mathrm{1}+{sin}\mathrm{2}{x}}\:=\int\:\frac{\mathrm{1}+{tan}^{\mathrm{2}} {x}}{\left(\mathrm{1}+{tanx}\right)^{\mathrm{2}} }\:{dx} \\ $$$$\overset{\mathrm{1}+{tanx}={u}} {=}\:\int\:\frac{{du}}{{u}}\:={lnu}+{c} \\ $$$$={ln}\left(\mathrm{1}+{tanx}\right)+{c}\:\checkmark \\ $$$$ \\ $$

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