# Question-142120

Question Number 142120 by bekzodjumayev last updated on 26/May/21
Commented by bekzodjumayev last updated on 26/May/21
$${Help} \\$$
Commented by Avijit007 last updated on 26/May/21
$${got}\:{it} \\$$
Answered by som(math1967) last updated on 26/May/21
$$\int\frac{{e}^{{x}} {dx}}{\mathrm{2}{cos}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}+\int\frac{\mathrm{2}\boldsymbol{{sin}}\frac{\boldsymbol{{x}}}{\mathrm{2}}\boldsymbol{{cos}}\frac{\boldsymbol{{x}}}{\mathrm{2}}\boldsymbol{{e}}^{\boldsymbol{{x}}} \boldsymbol{{dx}}}{\mathrm{2}\boldsymbol{{cos}}^{\mathrm{2}} \frac{\boldsymbol{{x}}}{\mathrm{2}}} \\$$$$\frac{\mathrm{1}}{\mathrm{2}}\int\boldsymbol{{sec}}^{\mathrm{2}} \frac{\boldsymbol{{x}}}{\mathrm{2}}\boldsymbol{{e}}^{\boldsymbol{{x}}} \boldsymbol{{dx}}\:+\int\boldsymbol{{tan}}\frac{\boldsymbol{{x}}}{\mathrm{2}}\boldsymbol{{e}}^{\boldsymbol{{x}}} \boldsymbol{{dx}} \\$$$$\frac{\mathrm{1}}{\mathrm{2}}\int\boldsymbol{{sec}}^{\mathrm{2}} \frac{\boldsymbol{{x}}}{\mathrm{2}}\boldsymbol{{e}}^{\boldsymbol{{x}}} \boldsymbol{{dx}}\:+\boldsymbol{{tan}}\frac{\boldsymbol{{x}}}{\mathrm{2}}\int\boldsymbol{{e}}^{\boldsymbol{{x}}} \boldsymbol{{dx}} \\$$$$\:\:\:\:\:\:\:\:\:\:\:−\int\left\{\frac{\boldsymbol{{d}}}{\boldsymbol{{dx}}}\boldsymbol{{tan}}\frac{\boldsymbol{{x}}}{\mathrm{2}}\int{e}^{\boldsymbol{{x}}} \boldsymbol{{dx}}\right\}\boldsymbol{{dx}} \\$$$$\frac{\mathrm{1}}{\mathrm{2}}\int\boldsymbol{{sec}}^{\mathrm{2}} \frac{\boldsymbol{{x}}}{\mathrm{2}}\boldsymbol{{e}}^{\boldsymbol{{x}}} \boldsymbol{{dx}}\:+\boldsymbol{{e}}^{\boldsymbol{{x}}} \boldsymbol{{tan}}\frac{\boldsymbol{{x}}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\int\boldsymbol{{sec}}^{\mathrm{2}} \frac{\boldsymbol{{x}}}{\mathrm{2}}\boldsymbol{{e}}^{\boldsymbol{{x}}} \boldsymbol{{dx}}\: \\$$$$\boldsymbol{{e}}^{\boldsymbol{{x}}} \boldsymbol{{tan}}\frac{\boldsymbol{{x}}}{\mathrm{2}}\:+\boldsymbol{{C}} \\$$
Commented by bekzodjumayev last updated on 26/May/21
$${Thank}\:{you} \\$$
Answered by Avijit007 last updated on 26/May/21
$${The}\:{given}\:{integral}\:{is}, \\$$$$\int\frac{\mathrm{1}+{sin}\:{x}}{\mathrm{1}+{cos}\:{x}}{e}^{{x}} .{dx} \\$$$$=\int\left[{e}^{{x}} \left\{\frac{\mathrm{1}}{\mathrm{1}+{cosx}}+\frac{{sinx}}{\mathrm{1}+{sinx}}\right\}\right].{dx} \\$$$$=\int\left[{e}^{{x}} \left\{\frac{\mathrm{1}}{\mathrm{2}{cos}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)}+\frac{\mathrm{2}{sin}\left(\frac{{x}}{\mathrm{2}}\right){cos}\left(\frac{{x}}{\mathrm{2}}\right)}{\mathrm{2}{cos}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)}\right].{dx}\right. \\$$$$=\int\left[{e}^{{x}} \left\{\frac{\mathrm{1}}{\mathrm{2}}{sec}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)+{tan}\left(\frac{{x}}{\mathrm{2}}\right)\right].{dx}\right. \\$$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Downarrow\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Downarrow \\$$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{f}\:'\left({x}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{f}\left({x}\right) \\$$$$\because\:{we}\:{know}\:{that},\:\int{e}^{{x}} \:\left\{{f}\:'\left({x}\right)+\:{f}\left({x}\right)\right\} \\$$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow{e}^{{x}} {f}\left({x}\right)+{c} \\$$$$=\:{e}^{{x}} .{tan}\left(\frac{{x}}{\mathrm{2}}\right)+{c}\:\:\:\:\:\:\left\{\mathcal{ANS}\right\}\:\: \\$$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{by}\:{Avijit} \\$$$$\\$$