Question-60637

Question Number 60637 by rajesh4661kumar@gamil.com last updated on 23/May/19

Commented by maxmathsup by imad last updated on 23/May/19

∫ ((sinx)/(1−sinx)) dx =−∫((1−sinx −1)/(1−sinx))dx =−x +∫ (dx/(1−sinx)) ∫ (dx/(1−sinx)) =_(tan((x/2))=t) ∫ (1/(1−((2t)/(1+t^2 )))) ((2dt)/(1+t^2 )) =2 ∫ (dt/(1+t^2 −2t)) =2 ∫ (dt/((t−1)^2 )) =((−2)/(t−1)) +c =(2/(1−tan((x/2)))) +c ⇒ ∫ ((sinx)/(1−sinx))dx =−x +(2/(1−tan((x/2)))) +c .

$$\int\:\:\:\:\frac{{sinx}}{\mathrm{1}−{sinx}}\:{dx}\:=−\int\frac{\mathrm{1}−{sinx}\:−\mathrm{1}}{\mathrm{1}−{sinx}}{dx}\:=−{x}\:+\int\:\:\frac{{dx}}{\mathrm{1}−{sinx}} \\ $$$$\int\:\:\frac{{dx}}{\mathrm{1}−{sinx}}\:=_{{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}} \:\:\:\:\:\:\int\:\:\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }}\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }\:=\mathrm{2}\:\int\:\:\:\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} −\mathrm{2}{t}}\:=\mathrm{2}\:\int\:\:\frac{{dt}}{\left({t}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\frac{−\mathrm{2}}{{t}−\mathrm{1}}\:+{c}\:=\frac{\mathrm{2}}{\mathrm{1}−{tan}\left(\frac{{x}}{\mathrm{2}}\right)}\:+{c}\:\Rightarrow \\ $$$$\int\:\:\frac{{sinx}}{\mathrm{1}−{sinx}}{dx}\:=−{x}\:+\frac{\mathrm{2}}{\mathrm{1}−{tan}\left(\frac{{x}}{\mathrm{2}}\right)}\:+{c}\:. \\ $$

Answered by tanmay last updated on 23/May/19

∫((sinx(1+sinx))/(cos^2 x))dx ∫((sinx)/(cos^2 x))dx+∫((sin^2 x)/(cos^2 x))dx ∫tanxsecxdx+∫(sec^2 x−1)dx ∫tanxsecx+∫sec^2 xdx−∫dx secx+tanx−x+c

$$\int\frac{{sinx}\left(\mathrm{1}+{sinx}\right)}{{cos}^{\mathrm{2}} {x}}{dx} \\ $$$$\int\frac{{sinx}}{{cos}^{\mathrm{2}} {x}}{dx}+\int\frac{{sin}^{\mathrm{2}} {x}}{{cos}^{\mathrm{2}} {x}}{dx} \\ $$$$\int{tanxsecxdx}+\int\left({sec}^{\mathrm{2}} {x}−\mathrm{1}\right){dx} \\ $$$$\int{tanxsecx}+\int{sec}^{\mathrm{2}} {xdx}−\int{dx} \\ $$$${secx}+{tanx}−{x}+{c} \\ $$$$ \\ $$

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Question-60637

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