Question-198901

Question Number 198901 by sonukgindia last updated on 25/Oct/23

Answered by witcher3 last updated on 25/Oct/23

∫(((cos((x/2))−sin((x/2)))^2 )/(6sin((x/2))cos((x/2))+10cos^2 ((x/2))e^x ))dx=g(x) =∫(((1−tg((x/2)))^2 )/(6tg((x/2))+10e^x ))dx =(1/(10))∫(((tg^2 ((x/2))−2tg((x/2))+1)e^(−x) )/((3/5)tg((x/2))e^(−x) +1)) f(x)=(3/5)tg((x/2))e^(−x) +1 f′(x)=(3/5)((1/2)(1+tg^2 ((x/2)))−tg((x/2)))e^(−x) =(3/(10))(tg^2 ((x/2))+1−2tg((x/2)))e^(−x) g(x)=(1/(10))∫.((10)/3).((f′)/f) (1/3)ln(∣(3/5)tg((x/2))e^(−x) +1∣)+c,c∈R

$$\int\frac{\left(\mathrm{cos}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)−\mathrm{sin}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)\right)^{\mathrm{2}} }{\mathrm{6sin}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)\mathrm{cos}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)+\mathrm{10cos}^{\mathrm{2}} \left(\frac{\mathrm{x}}{\mathrm{2}}\right)\mathrm{e}^{\mathrm{x}} }\mathrm{dx}=\mathrm{g}\left(\mathrm{x}\right) \\ $$$$=\int\frac{\left(\mathrm{1}−\mathrm{tg}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)\right)^{\mathrm{2}} }{\mathrm{6tg}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)+\mathrm{10e}^{\mathrm{x}} }\mathrm{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{10}}\int\frac{\left(\mathrm{tg}^{\mathrm{2}} \left(\frac{\mathrm{x}}{\mathrm{2}}\right)−\mathrm{2tg}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)+\mathrm{1}\right)\mathrm{e}^{−\mathrm{x}} }{\frac{\mathrm{3}}{\mathrm{5}}\mathrm{tg}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)\mathrm{e}^{−\mathrm{x}} +\mathrm{1}} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{3}}{\mathrm{5}}\mathrm{tg}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)\mathrm{e}^{−\mathrm{x}} +\mathrm{1} \\ $$$$\mathrm{f}'\left(\mathrm{x}\right)=\frac{\mathrm{3}}{\mathrm{5}}\left(\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\mathrm{tg}^{\mathrm{2}} \left(\frac{\mathrm{x}}{\mathrm{2}}\right)\right)−\mathrm{tg}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)\right)\mathrm{e}^{−\mathrm{x}} \\ $$$$=\frac{\mathrm{3}}{\mathrm{10}}\left(\mathrm{tg}^{\mathrm{2}} \left(\frac{\mathrm{x}}{\mathrm{2}}\right)+\mathrm{1}−\mathrm{2tg}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)\right)\mathrm{e}^{−\mathrm{x}} \\ $$$$\mathrm{g}\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{10}}\int.\frac{\mathrm{10}}{\mathrm{3}}.\frac{\mathrm{f}'}{\mathrm{f}} \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}\mathrm{ln}\left(\mid\frac{\mathrm{3}}{\mathrm{5}}\mathrm{tg}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)\mathrm{e}^{−\mathrm{x}} +\mathrm{1}\mid\right)+\mathrm{c},\mathrm{c}\in\mathbb{R} \\ $$$$ \\ $$

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