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Question Number 114037 by mohammad17 last updated on 16/Sep/20

$${\underset{0}{\int }}^{\frac{\pi }{4}}\frac{tan2x}{sin2x+cos2x}dx$$

Answered by Dwaipayan Shikari last updated on 16/Sep/20

$${\int }_0^{\frac{\pi }{4}}\frac{sin2x}{cos2x(sin2x+cos2x)}dx$$$${\int }_0^{\frac{\pi }{4}}\frac{1}{cos2x}-\frac{1}{sin2x+cos2x}dx$$$$={\int }_0^{\frac{\pi }{4}}sec2x-\frac{1}{\sqrt{2}}{\int }_0^{\frac{\pi }{4}}\frac{1}{cos(\frac{\pi }{4}-2x)}dx$$$$=\frac{1}{2}{\left[log(sec2x+tan2x)\right]}_0^{\frac{\pi }{4}}-\frac{1}{\sqrt{2}}{\int }_0^{\frac{\pi }{4}}sec(\frac{\pi }{4}-2x)dx$$$$\to \mathrm{\infty }$$$$Divergent$$

Answered by mathmax by abdo last updated on 16/Sep/20

$$\mathrm{I}={\int }_0^{\frac{\pi }{4}}\frac{\tan \left(2\mathrm{x}\right)}{\sin \left(2\mathrm{x}\right)+\cos \left(2\mathrm{x}\right)}\mathrm{dx}\Rightarrow \mathrm{I}{=}_{2\mathrm{x}=\mathrm{t}}\frac{1}{2}{\int }_0^{\frac{\pi }{2}}\frac{\tan \left(\mathrm{t}\right)}{\mathrm{sint}+\cos \left(\mathrm{t}\right)}\mathrm{dt}$$$${=}_{\tan \left(\frac{\mathrm{t}}{2}\right)=\mathrm{u}}\frac{1}{2}{\int }_0^1\frac{\frac{2\mathrm{u}}{1-{\mathrm{u}}^2}}{\frac{2\mathrm{u}}{1+{\mathrm{u}}^2}+\frac{1-{\mathrm{u}}^2}{1+{\mathrm{u}}^2}}\mathrm{du}=\frac{1}{2}{\int }_0^1\frac{2\mathrm{u}(1+{\mathrm{u}}^2)}{(2\mathrm{u}+1-{\mathrm{u}}^2)(1-{\mathrm{u}}^2)}\mathrm{du}$$$$={\int }_0^1\frac{\mathrm{u}({\mathrm{u}}^2+1)}{({\mathrm{u}}^2-2\mathrm{u}-1)({\mathrm{u}}^2-1)}\mathrm{du}\mathrm{but}\mathrm{at}\mathrm{v}\left(1\right)$$$$\frac{\mathrm{u}({\mathrm{u}}^2+1)}{({\mathrm{u}}^2-2\mathrm{u}-1)({\mathrm{u}}^2-1)}\sim \frac{2}{-2({\mathrm{u}}^2-1)}=\frac{1}{1-{\mathrm{u}}^2}\mathrm{and}{\int }_0^1\frac{\mathrm{du}}{1-{\mathrm{u}}^2}\mathrm{diverges}$$$$\Rightarrow \mathrm{the}\mathrm{intehral}\mathrm{I}\mathrm{is}\mathrm{dkvervent}...!$$

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