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dx-sin-x-1-cos-x-




Question Number 185985 by cortano1 last updated on 30/Jan/23
   ∫ (dx/( (√(sin x(1+cos x))))) =?
$$\:\:\:\int\:\frac{{dx}}{\:\sqrt{\mathrm{sin}\:{x}\left(\mathrm{1}+\mathrm{cos}\:{x}\right)}}\:=? \\ $$
Answered by ARUNG_Brandon_MBU last updated on 30/Jan/23
I=∫(dx/( (√(sinx(1+cosx)))))=∫(dx/(2(√(sin(x/2)cos^3 (x/2)))))    =∫(((1/2)sec^2 (x/2))/( (√(tan(x/2)))))dx=∫((d(tan(x/2)))/( (√(tan(x/2)))))=2(√(tan(x/2)))+C
$${I}=\int\frac{{dx}}{\:\sqrt{\mathrm{sin}{x}\left(\mathrm{1}+\mathrm{cos}{x}\right)}}=\int\frac{{dx}}{\mathrm{2}\sqrt{\mathrm{sin}\frac{{x}}{\mathrm{2}}\mathrm{cos}^{\mathrm{3}} \frac{{x}}{\mathrm{2}}}} \\ $$$$\:\:=\int\frac{\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sec}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}{\:\sqrt{\mathrm{tan}\frac{{x}}{\mathrm{2}}}}{dx}=\int\frac{{d}\left(\mathrm{tan}\frac{{x}}{\mathrm{2}}\right)}{\:\sqrt{\mathrm{tan}\frac{{x}}{\mathrm{2}}}}=\mathrm{2}\sqrt{\mathrm{tan}\frac{{x}}{\mathrm{2}}}+{C} \\ $$
Answered by Gamil last updated on 31/Jan/23
Commented by ARUNG_Brandon_MBU last updated on 31/Jan/23
(2/( (√(cscx+cotx))))=(2/( (√((1+cosx)/(sinx)))))  =2(√((2sin(x/2)cos(x/2))/(2cos^2 (x/2))))=2(√(tan(x/2)))
$$\frac{\mathrm{2}}{\:\sqrt{\mathrm{csc}{x}+\mathrm{cot}{x}}}=\frac{\mathrm{2}}{\:\sqrt{\frac{\mathrm{1}+\mathrm{cos}{x}}{\mathrm{sin}{x}}}} \\ $$$$=\mathrm{2}\sqrt{\frac{\mathrm{2sin}\frac{{x}}{\mathrm{2}}\mathrm{cos}\frac{{x}}{\mathrm{2}}}{\mathrm{2cos}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}}=\mathrm{2}\sqrt{\mathrm{tan}\frac{{x}}{\mathrm{2}}} \\ $$

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