Question Number 64448 by mathmax by abdo last updated on 18/Jul/19
$${calculate}\:{lim}_{{x}\rightarrow\pi} \:\:\int_{\frac{\pi}{\mathrm{2}}} ^{{x}} \:\:\:\frac{{dx}}{\mathrm{1}+{sinx}−{cosx}} \\ $$
Commented by mathmax by abdo last updated on 19/Jul/19
![let A(x) =∫_(π/2) ^x (dt/(1+sint−cost)) changement tan((t/2))=u vive A(x) =∫_1 ^(tan((x/2))) ((2du)/((1+u^2 )(1+((2u)/(1+u^2 ))−((1−u^2 )/(1+u^2 ))))) =∫_1 ^(tan((x/2))) ((2du)/(1+u^2 +2u−1+u^2 )) =∫_1 ^(tan((x/2))) ((2du)/(2u^2 +2u)) =∫_1 ^(tan((x/2))) (du/(u(u+1))) =∫_1 ^(tan((x/2))) {(1/u)−(1/(u+1))}du =[ln∣(u/(u+1))∣]_1 ^(tan((x/2))) =ln∣((tan((x/2)))/(tan((x/2))+1))∣−ln((1/2)) ⇒lim_(x→π) A(x) =0+ln(2) =ln(2).](Q64567.png)
$${let}\:{A}\left({x}\right)\:=\int_{\frac{\pi}{\mathrm{2}}} ^{{x}} \:\:\frac{{dt}}{\mathrm{1}+{sint}−{cost}}\:{changement}\:{tan}\left(\frac{{t}}{\mathrm{2}}\right)={u}\:{vive} \\ $$$${A}\left({x}\right)\:=\int_{\mathrm{1}} ^{{tan}\left(\frac{{x}}{\mathrm{2}}\right)} \:\:\:\frac{\mathrm{2}{du}}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\left(\mathrm{1}+\frac{\mathrm{2}{u}}{\mathrm{1}+{u}^{\mathrm{2}} }−\frac{\mathrm{1}−{u}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{2}} }\right)} \\ $$$$=\int_{\mathrm{1}} ^{{tan}\left(\frac{{x}}{\mathrm{2}}\right)} \:\frac{\mathrm{2}{du}}{\mathrm{1}+{u}^{\mathrm{2}} \:+\mathrm{2}{u}−\mathrm{1}+{u}^{\mathrm{2}} }\:=\int_{\mathrm{1}} ^{{tan}\left(\frac{{x}}{\mathrm{2}}\right)} \:\:\frac{\mathrm{2}{du}}{\mathrm{2}{u}^{\mathrm{2}} +\mathrm{2}{u}}\:=\int_{\mathrm{1}} ^{{tan}\left(\frac{{x}}{\mathrm{2}}\right)} \:\frac{{du}}{{u}\left({u}+\mathrm{1}\right)} \\ $$$$=\int_{\mathrm{1}} ^{{tan}\left(\frac{{x}}{\mathrm{2}}\right)} \left\{\frac{\mathrm{1}}{{u}}−\frac{\mathrm{1}}{{u}+\mathrm{1}}\right\}{du}\:=\left[{ln}\mid\frac{{u}}{{u}+\mathrm{1}}\mid\right]_{\mathrm{1}} ^{{tan}\left(\frac{{x}}{\mathrm{2}}\right)} \:={ln}\mid\frac{{tan}\left(\frac{{x}}{\mathrm{2}}\right)}{{tan}\left(\frac{{x}}{\mathrm{2}}\right)+\mathrm{1}}\mid−{ln}\left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\Rightarrow{lim}_{{x}\rightarrow\pi} \:{A}\left({x}\right)\:=\mathrm{0}+{ln}\left(\mathrm{2}\right)\:={ln}\left(\mathrm{2}\right). \\ $$$$ \\ $$
Answered by Tanmay chaudhury last updated on 18/Jul/19
$$\int\frac{{dx}}{\mathrm{2}{sin}\frac{{x}}{\mathrm{2}}{cos}\frac{{x}}{\mathrm{2}}+\mathrm{2}{sin}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}} \\ $$$$\int\frac{{sec}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}{dx}}{\mathrm{2}{tan}\frac{{x}}{\mathrm{2}}+\mathrm{2}{tan}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}{dx} \\ $$$$\int\frac{{d}\left({tan}\frac{{x}}{\mathrm{2}}\right)}{{tan}\frac{{x}}{\mathrm{2}}\left(\mathrm{1}+{tan}\frac{{x}}{\mathrm{2}}\right)} \\ $$$$\int\frac{{d}\left({tan}\frac{{x}}{\mathrm{2}}\right)}{{tan}\frac{{x}}{\mathrm{2}}}−\int\frac{{d}\left({tan}\frac{{x}}{\mathrm{2}}\right)}{\mathrm{1}+{tan}\frac{{x}}{\mathrm{2}}} \\ $$$$\mid{ln}\left(\frac{{tan}\frac{{x}}{\mathrm{2}}}{\mathrm{1}+{tan}\frac{{x}}{\mathrm{2}}}\right)\mid_{\frac{\pi}{\mathrm{2}}} ^{\pi} \\ $$$${ln}\mathrm{1}−{ln}\left(\frac{\mathrm{1}}{\mathrm{1}+\mathrm{1}}\right) \\ $$$$={ln}\mathrm{2} \\ $$