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Question Number 103825 by mohammad17 last updated on 17/Jul/20

$$\int \frac{dx}{{(1-sinx)}^2}?$$

Answered by ~blr237~ last updated on 17/Jul/20

$$\int \frac{1+2sinx+(1-{cos}^{2}x)}{{cos}^{4}x}dx$$$$\int [2(1+{tan}^{2}x)+2(1+{tan}^{2}x){tan}^{2}x+\frac{2sinx}{{\left(cosx\right)}^4}-\frac{1}{{cos}^{2}x})dx$$

Commented by mohammad17 last updated on 17/Jul/20

$$howthissir$$

Answered by mathmax by abdo last updated on 17/Jul/20

$$\mathrm{let}\mathrm{f}\left(\mathrm{a}\right)=\int \frac{\mathrm{dx}}{\mathrm{a}-\mathrm{sinx}}\mathrm{we}\mathrm{have}{\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)=-\int \frac{\mathrm{dx}}{{(\mathrm{a}-\mathrm{sinx})}^2}\Rightarrow$$$$\int \frac{\mathrm{dx}}{{(\mathrm{a}-\mathrm{sinx})}^2}=-{\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)\mathrm{and}\int \frac{\mathrm{dx}}{{(1-\mathrm{sinx})}^2}=-{\mathrm{f}}^{{}^{\prime }}\left(1\right)\mathrm{let}\mathrm{explicite}\mathrm{f}\left(\mathrm{a}\right)$$$$\mathrm{changement}\tan \left(\frac{\mathrm{x}}{2}\right)=\mathrm{t}\mathrm{give}\mathrm{f}\left(\mathrm{a}\right)=\int \frac{2\mathrm{dt}}{(1+{\mathrm{t}}^2)(\mathrm{a}-\frac{2\mathrm{t}}{1+{\mathrm{t}}^2})}$$$$=\int \frac{2\mathrm{dt}}{\mathrm{a}+{\mathrm{at}}^2-2\mathrm{t}}=\int \frac{2\mathrm{dt}}{{\mathrm{at}}^2-2\mathrm{t}+\mathrm{a}}$$$${\mathrm{\Delta }}^{{}^{\prime }}=1-{\mathrm{a}}^2\mathrm{let}\mathrm{take}\mathrm{a}>1\Rightarrow {\mathrm{\Delta }}^{{}^{\prime }}<0\Rightarrow \mathrm{f}\left(\mathrm{a}\right)=\frac{2}{\mathrm{a}}\int \frac{\mathrm{dt}}{{\mathrm{t}}^2-\frac{2\mathrm{t}}{\mathrm{a}}+1}$$$$=\frac{2}{\mathrm{a}}\int \frac{\mathrm{dt}}{{\mathrm{t}}^2-2\frac{\mathrm{t}}{\mathrm{a}}+\frac{1}{{\mathrm{a}}^2}+1-\frac{1}{{\mathrm{a}}^2}}=\frac{2}{\mathrm{a}}\int \frac{\mathrm{dt}}{{(\mathrm{t}-\frac{1}{\mathrm{a}})}^2+\frac{{\mathrm{a}}^2-1}{{\mathrm{a}}^2}}$$$${=}_{\mathrm{t}-\frac{1}{\mathrm{a}}=\frac{\sqrt{{\mathrm{a}}^2-1}}{\mathrm{a}}\mathrm{u}}\frac{2}{\mathrm{a}}\times \frac{{\mathrm{a}}^2}{{\mathrm{a}}^2-1}\int \frac{1}{1+{\mathrm{u}}^2}\times \frac{\sqrt{{\mathrm{a}}^2-1}}{\mathrm{a}}\mathrm{du}$$$$=\frac{2}{\sqrt{{\mathrm{a}}^2-1}}\int \frac{\mathrm{du}}{1+{\mathrm{u}}^2}=\frac{2}{\sqrt{{\mathrm{a}}^2-1}}\arctan \left(\frac{\mathrm{at}-1}{\sqrt{{\mathrm{a}}^2-1}}\right)=\frac{2}{\sqrt{{\mathrm{a}}^2-1}}\arctan \left(\frac{\mathrm{atan}\left(\frac{\mathrm{x}}{2}\right)-1}{\sqrt{{\mathrm{a}}^2-1}}\right)+\mathrm{c}$$$$\mathrm{rest}\mathrm{to}\mathrm{calculate}{\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)\mathrm{and}{\lim }_{\mathrm{a}\to 1}{\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)...\mathrm{be}\mathrm{continued}...$$

Answered by Dwaipayan Shikari last updated on 17/Jul/20

$$\int \frac{{(1+sinx)}^2}{{(1-{sin}^{2}x)}^2}dx=\int {({sec}^{2}x+secxtanx)}^{2}dx=\int {sec}^{2}x{(secx+tanx)}^{2}dx$$$$\int secx(secx+tanx)secx(secx+tanx)dx$$$$\int secxtdt\{takesecx+tanx=t\frac{dt}{dx}=secx(secx+tanx)$$$$\int \frac{t}{\sqrt{2t^2-t^4}}dt\{\frac{1+sinx}{cosx}=t\Rightarrow 1+{sin}^{2}x+2sinx=t^2(1-{sin}^{2}x)$$$$\int \frac{1}{\sqrt{2-t^2}}dt\{\Rightarrow (1+t^2){sin}^{2}x+2sinx+1-t^2=0$$$$\int \frac{\sqrt{2}cos\theta }{\sqrt{2-2{sin}^2\theta }}d\theta \{t=\sqrt{2}sin\theta \sqrt{2}cos\theta =\frac{dt}{d\theta }$$$$\int d\theta \{sinx=\frac{-2\pm \sqrt{4-4(1-t^4)}}{2}=t^2-1$$$$=\theta +C$$$$\{cosx=\sqrt{1-{(t^2-1)}^2}$$$$={sin}^{-1}\frac{t}{\sqrt{2}}+C$$$$={sin}^{-1}\left(\frac{secx+tanx}{\sqrt{2}}\right)+Constant$$$$$$

Commented by mohammad17 last updated on 17/Jul/20

$$thankyousir$$

Commented by Dwaipayan Shikari last updated on 17/Jul/20

$$Kindlycheckit$$

Commented by Dwaipayan Shikari last updated on 17/Jul/20

Commented by Dwaipayan Shikari last updated on 17/Jul/20

$$Alternatesolution$$

Answered by mathmax by abdo last updated on 17/Jul/20

$$\mathrm{let}\mathrm{try}\mathrm{another}\mathrm{way}$$$$\mathrm{I}=\int \frac{\mathrm{dx}}{{(1-\mathrm{sinx})}^2}\Rightarrow \mathrm{I}=\int \frac{\mathrm{dx}}{{(1-\cos (\frac{\pi }{2}-\mathrm{x}))}^2}=\int \frac{\mathrm{dx}}{{\left({2\sin }^2(\frac{\pi }{4}-\frac{\mathrm{x}}{2})\right)}^4}$$$$=\frac{1}{16}\int \frac{\mathrm{dx}}{{\sin }^4(\frac{\pi }{4}-\frac{\mathrm{x}}{2})}{=}_{\frac{\pi }{4}-\frac{\mathrm{x}}{2}=\mathrm{t}}\frac{1}{16}\int \frac{-2\mathrm{dt}}{{\sin }^4\mathrm{t}}=-\frac{1}{8}\int \frac{\mathrm{dt}}{{\sin }^4\mathrm{t}}$$$$\mathrm{changement}\tan \left(\frac{\mathrm{t}}{2}\right)=\mathrm{u}\mathrm{give}\int \frac{\mathrm{dt}}{{\sin }^4\mathrm{t}}=\int \frac{2\mathrm{du}}{(1+{\mathrm{u}}^2){\left(\frac{2\mathrm{u}}{1+{\mathrm{u}}^2}\right)}^4}$$$$=\int \frac{2{(1+\mathrm{u})}^3}{{16\mathrm{u}}^4}\mathrm{du}=\frac{1}{8}\int \frac{{\mathrm{u}}^3+{3\mathrm{u}}^2+3\mathrm{u}+1}{{\mathrm{u}}^4}\mathrm{du}$$$$=\frac{1}{8}\int (\frac{1}{\mathrm{u}}+\frac{3}{{\mathrm{u}}^2}+\frac{3}{{\mathrm{u}}^3}+\frac{1}{{\mathrm{u}}^4})\mathrm{du}$$$$=\frac{1}{8}\{\ln \mid \mathrm{u}\mid -\frac{3}{\mathrm{u}}+3(-\frac{1}{{2\mathrm{u}}^2})-\frac{1}{{3\mathrm{u}}^3}\}+\mathrm{C}$$$$\Rightarrow \mathrm{I}=-\frac{1}{64}\{\ln \mid \tan \left(\frac{\mathrm{t}}{2}\right)\mid -\frac{3}{\tan \left(\frac{\mathrm{t}}{2}\right)}-\frac{3}{{2\tan }^2\left(\frac{\mathrm{t}}{2}\right)}-\frac{1}{{3\tan }^3\left(\frac{\mathrm{t}}{2}\right)}\}+\mathrm{C}$$$$=-\frac{1}{64}\{\ln \mid \tan (\frac{\pi }{8}-\frac{\mathrm{x}}{4})\mid -\frac{3}{\tan (\frac{\pi }{8}-\frac{\mathrm{x}}{4})}-\frac{3}{{2\tan }^2(\frac{\pi }{8}-\frac{\mathrm{x}}{4})}-\frac{1}{{3\tan }^3(\frac{\pi }{8}-\frac{\mathrm{x}}{4})}\}+\mathrm{C}$$

Commented by mathmax by abdo last updated on 17/Jul/20

$$\mathrm{error}\mathrm{of}\mathrm{typo}\mathrm{I}=\int \frac{\mathrm{dx}}{{\left({2\sin }^2(\frac{\pi }{4}-\frac{\mathrm{x}}{2})\right)}^2}=...$$

Commented by mathmax by abdo last updated on 17/Jul/20

$$\mathrm{sorry}\mathrm{change}64\mathrm{by}32...$$

Commented by mohammad17 last updated on 17/Jul/20

$$oksirthankyou$$

Answered by PRITHWISH SEN 2 last updated on 17/Jul/20

$$\int \frac{\mathrm{dx}}{{(1-\frac{2\tan \frac{\mathrm{x}}{2}}{1+\tan {}^2\frac{\mathrm{x}}{2}})}^2}=\int \frac{{\sec }^2\frac{\mathrm{x}}{2}}{{(1-\tan \frac{\mathrm{x}}{2})}^4}(1+\tan {}^2\frac{\mathrm{x}}{2})\mathrm{dx}$$$$\tan \frac{\mathrm{x}}{2}=\mathrm{t}\Rightarrow \sec {}^2\frac{\mathrm{x}}{2}\mathrm{dx}=2\mathrm{dt}$$$$=2\int \frac{(1+{\mathrm{t}}^2)\mathrm{dt}}{{(1-\mathrm{t})}^4}=\int \frac{2\mathrm{dt}}{{(1-\mathrm{t})}^2}-\int \frac{4(1-\mathrm{t})\mathrm{dt}}{{(1-\mathrm{t})}^4}+4\int \frac{\mathrm{dt}}{{(1-\mathrm{t})}^4}$$$$=\frac{2}{(1-\mathrm{t})}+\frac{2}{{(1-\mathrm{t})}^2}-\frac{4}{3{(1-\mathrm{t})}^3}+\mathrm{C}$$$$=\frac{2}{(1-\tan \frac{\mathrm{x}}{2})}+\frac{2}{{(1-\tan \frac{\mathrm{x}}{2})}^2}-\frac{4}{3{(1-\tan \frac{\mathrm{x}}{2})}^3}+\mathrm{C}$$$$\mathrm{please}\mathrm{check}$$

Commented by mohammad17 last updated on 17/Jul/20

$$thanyousir$$

Answered by Ar Brandon last updated on 19/Jul/20

$$\mathcal{I}=\int \frac{\mathrm{dx}}{{(1-\mathrm{sinx})}^2}=\int \frac{{(1+\mathrm{sinx})}^2}{{(1-\mathrm{sinx})}^2{(1+\mathrm{sinx})}^2}\mathrm{dx}$$$$=\int \frac{1+2\mathrm{sinx}+{\sin }^2\mathrm{x}}{{(1-{\sin }^2\mathrm{x})}^2}\mathrm{dx}=\int \frac{{\cos }^2\mathrm{x}+2\mathrm{sinx}+{2\sin }^2\mathrm{x}}{{\cos }^4\mathrm{x}}\mathrm{dx}$$$$=\int {\sec }^2\mathrm{xdx}+2\int \frac{\mathrm{sinx}}{{\cos }^4\mathrm{x}}\mathrm{dx}+2\int {\tan }^2{\mathrm{xsec}}^2\mathrm{xdx}$$$$=\mathrm{tanx}+\frac{{2\sec }^3\mathrm{x}}{3}+\frac{{2\tan }^3\mathrm{x}}{3}+\mathcal{C}$$

Answered by Her_Majesty last updated on 07/Aug/20

$$\int \frac{dx}{{(1-\sin x)}^2}$$$$\mathrm{simply}{\mathrm{Weyerstrass}}^{\prime }\mathrm{Substitution}$$$$t=\tan \frac{x}{2}$$$$\mathrm{leads}\mathrm{to}$$$$2\int \frac{t^2+1}{{(t-1)}^4}dt=-\frac{2(3t^2-3t+2)}{3{(t-1)}^3}=...$$

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