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Question-68046




Question Number 68046 by mhmd last updated on 03/Sep/19
Answered by mind is power last updated on 03/Sep/19
((d(e^((sin(2x))/2) ))/dx)=cos(2x)e^(sin(x)cos(x))   ∫((e^(sin(x)cos(x)) cos(2x))/(1+e^(sin(2x)) ))dx  =∫((d(e^((sin(2x))/2) ))/(1+e^(2((sin(2x))/2)) ))=∫((d(e^((sin(2x))/2) ))/(1+(e^((sin(2x))/2) )^2 ))=tan^(−1) (e^((sin(2x))/2) )+c
$$\frac{{d}\left({e}^{\frac{{sin}\left(\mathrm{2}{x}\right)}{\mathrm{2}}} \right)}{{dx}}={cos}\left(\mathrm{2}{x}\right){e}^{{sin}\left({x}\right){cos}\left({x}\right)} \\ $$$$\int\frac{{e}^{{sin}\left({x}\right){cos}\left({x}\right)} {cos}\left(\mathrm{2}{x}\right)}{\mathrm{1}+{e}^{{sin}\left(\mathrm{2}{x}\right)} }{dx} \\ $$$$=\int\frac{{d}\left({e}^{\frac{{sin}\left(\mathrm{2}{x}\right)}{\mathrm{2}}} \right)}{\mathrm{1}+{e}^{\mathrm{2}\frac{{sin}\left(\mathrm{2}{x}\right)}{\mathrm{2}}} }=\int\frac{{d}\left({e}^{\frac{{sin}\left(\mathrm{2}{x}\right)}{\mathrm{2}}} \right)}{\mathrm{1}+\left({e}^{\frac{{sin}\left(\mathrm{2}{x}\right)}{\mathrm{2}}} \right)^{\mathrm{2}} }=\mathrm{tan}^{−\mathrm{1}} \left({e}^{\frac{{sin}\left(\mathrm{2}{x}\right)}{\mathrm{2}}} \right)+{c} \\ $$

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