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Question Number 106709 by bemath last updated on 06/Aug/20

$$$$$$\int \frac{\sin {}^2\mathrm{x}\mathrm{dx}}{1-\sin \mathrm{x}\cos \mathrm{x}}?$$

Answered by john santu last updated on 06/Aug/20

$$\stackrel{@\mathrm{JS}@}{}$$$$\mathrm{I}=\int \frac{\frac{1}{2}-\frac{1}{2}\cos 2\mathrm{x}}{1-\frac{\sin 2\mathrm{x}}{2}}\mathrm{dx}=\int \frac{1-\cos 2\mathrm{x}}{2-\sin 2\mathrm{x}}\mathrm{dx}$$$$\mathrm{I}=\int \frac{\mathrm{dx}}{2-\sin 2\mathrm{x}}-\int \frac{\cos 2\mathrm{x}}{2-\sin 2\mathrm{x}}\mathrm{dx}$$$$\mathrm{set}{\mathrm{I}}_1=\int \frac{\mathrm{dx}}{2-\sin 2\mathrm{x}}[\mathrm{let}\tan \mathrm{x}=\mathrm{♭}]$$$${\mathrm{I}}_1=\int \frac{\left(\frac{1}{{\mathrm{♭}}^2+1}\right)\mathrm{d}\mathrm{♭}}{2-\left(\frac{2\mathrm{♭}}{{\mathrm{♭}}^2+1}\right)}=\int \frac{\mathrm{d}\mathrm{♭}}{2{\mathrm{♭}}^2-2\mathrm{♭}+2}$$$${\mathrm{I}}_1=\frac{1}{2}\int \frac{\mathrm{d}\mathrm{♭}}{{(\mathrm{♭}-\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}$$$${\mathrm{I}}_2=\int \frac{\cos 2\mathrm{x}\mathrm{dx}}{2-\sin 2\mathrm{x}}=-\frac{1}{2}\int \frac{\mathrm{d}(2-\sin 2\mathrm{x})}{2-\sin 2\mathrm{x}}$$$${\mathrm{I}}_1=-\frac{1}{2}\ln \mid 2-\sin 2\mathrm{x}\mid$$$$\therefore \mathrm{I}={\mathrm{I}}_1-{\mathrm{I}}_2$$

Answered by Dwaipayan Shikari last updated on 06/Aug/20

$$\int \frac{\frac{1-cos2x}{2}}{1-\frac{1}{2}sin2x}dx$$$$\int \frac{1-cos2x}{2-sin2x}dx=\int \frac{1}{2-sin2x}+\frac{1}{2}\int \frac{-2cos2x}{2-sin2x}$$$$=\int \frac{1}{2-\frac{2t}{1+t^2}}.\frac{1}{1+t^2}+\frac{1}{2}log(2-sin2x)(t=tanx)$$$$=\int \frac{1}{1+t^2-t}dt+\frac{1}{2}log(2-sin2x)$$$$=\frac{1}{2}\int \frac{1}{{(t-\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}+\frac{1}{2}log(2-sin2x)$$$$=\frac{1}{\sqrt{3}}{tan}^{-1}\frac{2t-1}{\sqrt{3}}+\frac{1}{2}log(2-sin2x)+C$$

Answered by mathmax by abdo last updated on 07/Aug/20

$$\mathrm{I}=\int \frac{{\sin }^2\mathrm{x}}{1-\mathrm{sinx}\mathrm{cosx}}\mathrm{dx}\Rightarrow \mathrm{I}=\int \frac{1-\cos \left(2\mathrm{x}\right)}{2(1-\frac{1}{2}\sin \left(2\mathrm{x}\right))}\mathrm{dx}$$$$=\int \frac{1-\cos \left(2\mathrm{x}\right)}{2-\sin \left(2\mathrm{x}\right)}\mathrm{dx}{=}_{\mathrm{tanx}=\mathrm{t}}\int \frac{1-\frac{1-{\mathrm{t}}^2}{1+{\mathrm{t}}^2}}{2-\frac{2\mathrm{t}}{1+{\mathrm{t}}^2}}\times \frac{\mathrm{dt}}{1+{\mathrm{t}}^2}$$$$=\int \frac{1+{\mathrm{t}}^2-1+{\mathrm{t}}^2}{2+{2\mathrm{t}}^2-2\mathrm{t}}\frac{\mathrm{dt}}{1+{\mathrm{t}}^2}=\frac{1}{2}\int \frac{{2\mathrm{t}}^2}{(1+{\mathrm{t}}^2)({\mathrm{t}}^2-\mathrm{t}+1)}\mathrm{dt}$$$$=\int \frac{{\mathrm{t}}^2}{({\mathrm{t}}^2+1)({\mathrm{t}}^2-\mathrm{t}+1)}\mathrm{dt}=-\int \mathrm{t}(\frac{1}{{\mathrm{t}}^2+1}-\frac{1}{{\mathrm{t}}^2-\mathrm{t}+1})\mathrm{dt}$$$$=\int (\frac{\mathrm{t}}{{\mathrm{t}}^2-\mathrm{t}+1}-\frac{\mathrm{t}}{{\mathrm{t}}^2+1})\mathrm{dt}=\frac{1}{2}\int \frac{2\mathrm{t}-1-1}{{\mathrm{t}}^2-\mathrm{t}+1}\mathrm{dt}-\frac{1}{2}\int \frac{2\mathrm{t}}{{\mathrm{t}}^2+1}\mathrm{dt}$$$$=\frac{1}{2}\ln ({\mathrm{t}}^2-\mathrm{t}+1)-\frac{1}{2}\ln ({\mathrm{t}}^2+1)-\frac{1}{2}\int \frac{\mathrm{dt}}{{\mathrm{t}}^2-\mathrm{t}+1}$$$$\mathrm{we}\mathrm{have}\int \frac{\mathrm{dt}}{{\mathrm{t}}^2-\mathrm{t}+1}=\int \frac{\mathrm{dt}}{{(\mathrm{t}-\frac{1}{2})}^2+\frac{3}{4}}{=}_{\mathrm{t}-\frac{1}{2}=\frac{\sqrt{3}}{2}\mathrm{u}}\frac{4}{3}\int \frac{1}{1+{\mathrm{u}}^2}.\frac{\sqrt{3}}{2}\mathrm{du}$$$$=\frac{2}{\sqrt{3}}\arctan \left(\mathrm{u}\right)+\mathrm{c}=\frac{2}{\sqrt{3}}\arctan \left(\frac{2\mathrm{t}-1}{\sqrt{3}}\right)\Rightarrow$$$$\mathrm{I}=\frac{1}{2}\ln ({\mathrm{t}}^2-\mathrm{t}+1)-\frac{1}{2}\ln ({\mathrm{t}}^2+1)-\frac{1}{\sqrt{3}}\arctan \left(\frac{2\mathrm{t}-1}{\sqrt{3}}\right)+\mathrm{C}$$$$\mathrm{I}=\frac{1}{2}\ln \left(\frac{{\tan }^2\mathrm{x}-\mathrm{tanx}+1}{{\tan }^2\mathrm{x}+1}\right)-\frac{1}{\sqrt{3}}\arctan \left(\frac{2\mathrm{tanx}-1}{\sqrt{3}}\right)+\mathrm{C}$$$$$$

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