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tan-1-sinx-1-sinx-dx-

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Question Number 29043 by yesaditya22@gmail.com last updated on 03/Feb/18
∫tan^− (1−sinx/1+sinx) dx
$$\int\mathrm{tan}^{−} \left(\mathrm{1}−\mathrm{sinx}/\mathrm{1}+\mathrm{sinx}\right)\:\mathrm{dx} \\ $$
Commented by abdo imad last updated on 03/Feb/18
let put I= ∫ arctan(((1−sinx)/(1+sinx)))dx (arctan=tan^(−1) )we have ((1−sinx)/(1+sinx))=((1 −cos((π/2)−x))/(1+cos((π/2)−x)))=((2sin^2 ((π/4)−(x/2)))/(2 cos^2 ((π/4) −(x/2))))=tan^2 ((π/4)−(x/2)) I = ∫ arctan(tan^2 ((π/4)−(x/2)))dx let use the ch. tan^2 ((π/4)−(x/2))=u ⇔tan((π/4)−(x/2)) =(√u) and (π/4)−(x/2)=arctan((√u))⇔ x=−2arctan((√u))⇒ dx=−2((1/(2(√u)))/(1+u))= ((−1)/( (√u)(1+u))) so I=−∫ ((artanu)/( (√u)(1+u)))du ((√u)=t) I=− ∫ ((arctan(t^2 ))/(t( 1+t^2 )))2t dt=−2∫ ((arctan(t^2 ))/(1+t^2 ))dt +λ and the integral ∫ ((arctan(t^2 ))/(1+t^2 ))dt be calculated by the parametric function f(x)= ∫ ((arctan(x(1+t^2 )))/(1+t^2 ))dt by derivation under ∫....
$${let}\:{put}\:{I}=\:\int\:{arctan}\left(\frac{\mathrm{1}−{sinx}}{\mathrm{1}+{sinx}}\right){dx}\:\:\:\left({arctan}={tan}^{−\mathrm{1}} \right){we}\:{have} \\ $$$$\frac{\mathrm{1}−{sinx}}{\mathrm{1}+{sinx}}=\frac{\mathrm{1}\:−{cos}\left(\frac{\pi}{\mathrm{2}}−{x}\right)}{\mathrm{1}+{cos}\left(\frac{\pi}{\mathrm{2}}−{x}\right)}=\frac{\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{4}}−\frac{{x}}{\mathrm{2}}\right)}{\mathrm{2}\:{cos}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{4}}\:−\frac{{x}}{\mathrm{2}}\right)}={tan}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{4}}−\frac{{x}}{\mathrm{2}}\right) \\ $$$${I}\:=\:\int\:{arctan}\left({tan}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{4}}−\frac{{x}}{\mathrm{2}}\right)\right){dx}\:\:{let}\:{use}\:{the}\:{ch}. \\ $$$${tan}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{4}}−\frac{{x}}{\mathrm{2}}\right)={u}\:\Leftrightarrow{tan}\left(\frac{\pi}{\mathrm{4}}−\frac{{x}}{\mathrm{2}}\right)\:=\sqrt{{u}}\:{and} \\ $$$$\frac{\pi}{\mathrm{4}}−\frac{{x}}{\mathrm{2}}={arctan}\left(\sqrt{{u}}\right)\Leftrightarrow\:{x}=−\mathrm{2}{arctan}\left(\sqrt{{u}}\right)\Rightarrow \\ $$$${dx}=−\mathrm{2}\frac{\frac{\mathrm{1}}{\mathrm{2}\sqrt{{u}}}}{\mathrm{1}+{u}}=\:\frac{−\mathrm{1}}{\:\sqrt{{u}}\left(\mathrm{1}+{u}\right)}\:{so} \\ $$$${I}=−\int\:\frac{{artanu}}{\:\sqrt{{u}}\left(\mathrm{1}+{u}\right)}{du}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\sqrt{{u}}={t}\right) \\ $$$${I}=−\:\int\:\:\:\frac{{arctan}\left({t}^{\mathrm{2}} \right)}{{t}\left(\:\mathrm{1}+{t}^{\mathrm{2}} \right)}\mathrm{2}{t}\:{dt}=−\mathrm{2}\int\:\:\frac{{arctan}\left({t}^{\mathrm{2}} \right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\:+\lambda\:{and}\:{the} \\ $$$${integral}\:\int\:\:\frac{{arctan}\left({t}^{\mathrm{2}} \right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\:{be}\:{calculated}\:{by}\:{the}\:{parametric} \\ $$$${function}\:{f}\left({x}\right)=\:\int\:\:\:\frac{{arctan}\left({x}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\:{by}\:{derivation}\:{under} \\ $$$$\int…. \\ $$$$ \\ $$

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