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Question Number 212239 by Spillover last updated on 07/Oct/24

Answered by Ghisom last updated on 08/Oct/24

∫((e^(−x/2) (√(1−sin x)))/(1+cos x))dx= [t=tan (x/2)] =∫((e^(−arctan t) (t−1))/( (√(t^2 +1))))dt= =∫ ((e^(−arctan t) t)/( (√(t^2 +1))))dt−∫ (e^(−arctan t) /( (√(t^2 +1))))dt= [by parts] =e^(−arctan t) (√(t^2 +1))+∫ (e^(−arctan t) /( (√(t^2 +1))))dt−∫ (e^(−arctan t) /( (√(t^2 +1))))dt= =e^(−arctan t) (√(t^2 +1))= =(e^(−x/2) /(cos (x/2)))+C

$$\int\frac{\mathrm{e}^{−{x}/\mathrm{2}} \sqrt{\mathrm{1}−\mathrm{sin}\:{x}}}{\mathrm{1}+\mathrm{cos}\:{x}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\right] \\ $$$$=\int\frac{\mathrm{e}^{−\mathrm{arctan}\:{t}} \left({t}−\mathrm{1}\right)}{\:\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}}{dt}= \\ $$$$=\int\:\frac{\mathrm{e}^{−\mathrm{arctan}\:{t}} {t}}{\:\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}}{dt}−\int\:\frac{\mathrm{e}^{−\mathrm{arctan}\:{t}} }{\:\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}}{dt}= \\ $$$$\:\:\:\:\:\left[\mathrm{by}\:\mathrm{parts}\right] \\ $$$$=\mathrm{e}^{−\mathrm{arctan}\:{t}} \sqrt{{t}^{\mathrm{2}} +\mathrm{1}}+\int\:\frac{\mathrm{e}^{−\mathrm{arctan}\:{t}} }{\:\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}}{dt}−\int\:\frac{\mathrm{e}^{−\mathrm{arctan}\:{t}} }{\:\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}}{dt}= \\ $$$$=\mathrm{e}^{−\mathrm{arctan}\:{t}} \sqrt{{t}^{\mathrm{2}} +\mathrm{1}}= \\ $$$$=\frac{\mathrm{e}^{−{x}/\mathrm{2}} }{\mathrm{cos}\:\frac{{x}}{\mathrm{2}}}+{C} \\ $$

Commented by Frix last updated on 08/Oct/24

Nice!

$$\mathrm{Nice}! \\ $$

Commented by Spillover last updated on 08/Oct/24

great.thanks

$${great}.{thanks} \\ $$

Commented by MathematicalUser2357 last updated on 12/Oct/24

No, the real answer is ((e^(−x/2) (√(1−sin x)) sec^2 (x/2))/(tan (x/2)+1))+C.

$$\mathrm{No},\:\mathrm{the}\:\mathrm{real}\:\mathrm{answer}\:\mathrm{is}\:\frac{{e}^{−{x}/\mathrm{2}} \sqrt{\mathrm{1}−\mathrm{sin}\:{x}}\:\mathrm{sec}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}{\mathrm{tan}\:\frac{{x}}{\mathrm{2}}+\mathrm{1}}+{C}. \\ $$

Commented by Ghisom last updated on 13/Oct/24

it′s the same, just transform it to see

$$\mathrm{it}'\mathrm{s}\:\mathrm{the}\:\mathrm{same},\:\mathrm{just}\:\mathrm{transform}\:\mathrm{it}\:\mathrm{to}\:\mathrm{see} \\ $$

Answered by MathematicalUser2357 last updated on 12/Oct/24

((e^(−x/2) (√(1−sin x)) sec^2 (x/2))/(tan (x/2)−1))+C

$$\frac{{e}^{−{x}/\mathrm{2}} \sqrt{\mathrm{1}−\mathrm{sin}\:{x}}\:\mathrm{sec}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}{\mathrm{tan}\:\frac{{x}}{\mathrm{2}}−\mathrm{1}}+{C} \\ $$

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