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Question Number 120254 by bramlexs22 last updated on 30/Oct/20

$$\int \frac{dx}{1+\cos \theta .\cos x}?$$

Answered by TANMAY PANACEA last updated on 30/Oct/20

$$\int \frac{dx}{cos\theta (sec\theta +cosx)}$$$$\frac{1}{cos\theta }\int \frac{dx}{sec\theta +\frac{1-{tan}^2\frac{x}{2}}{1+{tan}^2\frac{x}{2}}}$$$$\frac{1}{cos\theta }\int \frac{{sec}^2\frac{x}{2}}{sec\theta +sec\theta {tan}^2\frac{x}{2}+1-{tan}^2\frac{x}{2}}$$$$\frac{1}{cos\theta }\int \frac{d\left(tan\frac{x}{2}\right)\times 2}{(1+sec\theta )+(sec\theta -1){tan}^2\frac{x}{2}}$$$$\frac{2}{cos\theta }\int \frac{dk}{(1+sec\theta )+(sec\theta -1)k^2}$$$$\frac{2}{cos\theta (sec\theta -1)}\int \frac{dk}{k^2+\frac{1+sec\theta }{sec\theta -1}}$$$$\frac{2}{1-cos\theta }\times \frac{1}{\sqrt{\frac{1+sec\theta }{sec\theta -1}}}{tan}^{-1}\left(\frac{k}{\sqrt{\frac{sec\theta +1}{sec\theta -1}}}\right)$$

Answered by bemath last updated on 30/Oct/20

$$let\cos \theta =k\Rightarrow \int \frac{dx}{1+k\cos x}$$$$let\tan \left(\frac{x}{2}\right)=u\Rightarrow \frac{1}{2}\sec {}^2\left(\frac{x}{2}\right)dx=du$$$$dx=2\cos {}^2\left(\frac{x}{2}\right)du=\frac{2du}{1+u^2}$$$$I=\int \frac{2du}{(1+u^2)(1+\frac{k(1-u^2)}{1+u^2})}$$$$I=\int \frac{2du}{(1+k)+(1-k)u^2}=\int \frac{2du}{{\left(\sqrt{1+k}\right)}^2+u^2{\left(\sqrt{1-k}\right)}^2}$$$$lettingu\sqrt{1-k}=\sqrt{1+k}\tan w$$$$I=\int \frac{2\left(\sqrt{\frac{1+k}{1-k}}\right)\sec {}^2\left(w\right)dw}{(1+k)\sec {}^2\left(w\right)}$$$$I=\frac{2}{\sqrt{1-k^2}}w+c=\frac{2}{\sqrt{1-k^2}}\mathrm{arc}\tan \left(u\sqrt{\frac{1-k}{1+k}}\right)+c$$$$=\frac{2}{\sqrt{1-\cos {}^2\theta }}\mathrm{arc}\tan \left(\tan \left(\frac{x}{2}\right)\sqrt{\frac{1-\cos \theta }{1+\cos \theta }}\right)+c$$$$=2\mathrm{cosec}\theta \mathrm{arc}\tan \left(\tan \left(\frac{x}{2}\right)\sqrt{\frac{1-\cos \theta }{1+\cos \theta }}\right)+c$$$$$$

Answered by mathmax by abdo last updated on 30/Oct/20

$$\mathrm{I}=\int \frac{\mathrm{dx}}{1+\cos \theta \mathrm{cosx}}=\int \frac{\mathrm{dx}}{1+\mathrm{a}\mathrm{cosx}}(\mathrm{a}=\cos \theta \Rightarrow \mid \mathrm{a}\mid ⩽1)$$$${=}_{\tan \left(\frac{\mathrm{x}}{2}\right)=\mathrm{t}}\int \frac{2\mathrm{dt}}{(1+{\mathrm{t}}^2)(1+\mathrm{a}\frac{1-{\mathrm{t}}^2}{1+{\mathrm{t}}^2})}=\int \frac{2\mathrm{dt}}{1+{\mathrm{t}}^2+\mathrm{a}-{\mathrm{at}}^2}$$$$=\int \frac{2\mathrm{dt}}{(1-\mathrm{a}){\mathrm{t}}^2+1+\mathrm{a}}=\frac{2}{1-\mathrm{a}}\int \frac{\mathrm{dt}}{{\mathrm{t}}^2+\frac{1+\mathrm{a}}{1-\mathrm{a}}}$$$${=}_{\mathrm{t}=\sqrt{\frac{1+\mathrm{a}}{1-\mathrm{a}}}\mathrm{z}}\frac{2}{1-\mathrm{a}}.\frac{1-\mathrm{a}}{1+\mathrm{a}}\int \frac{1}{{\mathrm{z}}^2+1}\frac{\sqrt{1+\mathrm{a}}}{\sqrt{1-\mathrm{a}}}\mathrm{dz}$$$$=\frac{2}{\sqrt{1-{\mathrm{a}}^2}}\arctan \left(\mathrm{z}\right)+\mathrm{c}=\frac{2}{\mid \sin \theta \mid }\arctan \left(\sqrt{\frac{1-\cos \theta }{1+\cos \theta }}\tan \left(\frac{\mathrm{x}}{2}\right)\right)+\mathrm{C}$$$$\mathrm{I}=\frac{2}{\mid \sin \theta \mid }\arctan (\mid \tan \left(\frac{\theta }{2}\right)\mid \tan \left(\frac{\mathrm{x}}{2}\right))+\mathrm{C}$$

Answered by Dwaipayan Shikari last updated on 30/Oct/20

$$\int \frac{dx}{1+\mathrm{\Gamma }cosx}\mathrm{\Gamma }=cos\theta$$$$=2\int \frac{dt}{1+\frac{\mathrm{\Gamma }-\mathrm{\Gamma }{t}^2}{1+t^2}}.\frac{1}{1+t^2}$$$$=2\int \frac{dt}{t^2(1-\mathrm{\Gamma })+(\mathrm{\Gamma }+1)}=\frac{2}{1-\mathrm{\Gamma }}\int \frac{1}{t^2+{\left(\sqrt{\frac{1+\mathrm{\Gamma }}{1-\mathrm{\Gamma }}}\right)}^2}dt$$$$=\frac{2}{\sqrt{1-{\mathrm{\Gamma }}^2}}{tan}^{-1}t\sqrt{\frac{1-\mathrm{\Gamma }}{1+\mathrm{\Gamma }}}+C$$$$=\frac{2}{\sqrt{1-{cos}^2\theta }}{tan}^{-1}tan\frac{x}{2}.\sqrt{\frac{1-cos\theta }{1+cos\theta }}+C$$$$\frac{2}{\mid sin\theta \mid }{tan}^{-1}(\left(tan\frac{x}{2}\right)\mid \left(tan\frac{\theta }{2}\right)\mid )+C$$

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