Question Number 44993 by arvinddayama01@gmail.com last updated on 07/Oct/18
$$\int\frac{\mathrm{x}}{\mathrm{sin}\:\mathrm{x}}\mathrm{dx}=? \\ $$
Commented by maxmathsup by imad last updated on 07/Oct/18
![let I (t)=∫_0 ^t (x/(sinx))dx changement tan((x/2))=u give I(t) = ∫_0 ^(tan((t/2))) ((2arctan(u))/((2u)/(1+u^2 ))) ((2du)/(1+u^2 )) =2∫_0 ^(tan((t/2))) ((arctan(u))/u)du let introduce the parametric function f(x) =∫_0 ^(tan((t/2))) ((arctan(xu))/u)du ⇒f^′ (x) =∫_0 ^(tan((t/2))) (1/(1+x^2 u^2 ))du =_(xu =α) ∫_0 ^(xtan((t/2))) (1/(1+α^2 )) (dα/x) =(1/x) [arctan(α)]_0 ^(xtan((t/2))) =(1/x) arctan(x tan((t/2))) ⇒ f(x) =∫_0 ^x ((arctan(u tan((t/2))))/u) du +c I(t) =2 f(1) =2 ∫_0 ^1 ((arctan(utan((t/2))))/u)du +c ...be continued...](https://www.tinkutara.com/question/Q45015.png)
$${let}\:{I}\:\left({t}\right)=\int_{\mathrm{0}} ^{{t}} \:\frac{{x}}{{sinx}}{dx}\:{changement}\:{tan}\left(\frac{{x}}{\mathrm{2}}\right)={u}\:{give} \\ $$$${I}\left({t}\right)\:=\:\int_{\mathrm{0}} ^{{tan}\left(\frac{{t}}{\mathrm{2}}\right)} \:\:\:\frac{\mathrm{2}{arctan}\left({u}\right)}{\frac{\mathrm{2}{u}}{\mathrm{1}+{u}^{\mathrm{2}} }}\:\frac{\mathrm{2}{du}}{\mathrm{1}+{u}^{\mathrm{2}} }\:=\mathrm{2}\int_{\mathrm{0}} ^{{tan}\left(\frac{{t}}{\mathrm{2}}\right)} \:\:\:\frac{{arctan}\left({u}\right)}{{u}}{du}\:{let}\:{introduce}\:{the}\:{parametric} \\ $$$${function}\:\:{f}\left({x}\right)\:=\int_{\mathrm{0}} ^{{tan}\left(\frac{{t}}{\mathrm{2}}\right)} \:\frac{{arctan}\left({xu}\right)}{{u}}{du}\:\Rightarrow{f}^{'} \left({x}\right)\:=\int_{\mathrm{0}} ^{{tan}\left(\frac{{t}}{\mathrm{2}}\right)} \:\:\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} {u}^{\mathrm{2}} }{du} \\ $$$$=_{{xu}\:=\alpha} \:\:\:\int_{\mathrm{0}} ^{{xtan}\left(\frac{{t}}{\mathrm{2}}\right)} \:\:\:\frac{\mathrm{1}}{\mathrm{1}+\alpha^{\mathrm{2}} }\:\frac{{d}\alpha}{{x}}\:=\frac{\mathrm{1}}{{x}}\:\left[{arctan}\left(\alpha\right)\right]_{\mathrm{0}} ^{{xtan}\left(\frac{{t}}{\mathrm{2}}\right)} \\ $$$$=\frac{\mathrm{1}}{{x}}\:{arctan}\left({x}\:{tan}\left(\frac{{t}}{\mathrm{2}}\right)\right)\:\Rightarrow\:{f}\left({x}\right)\:=\int_{\mathrm{0}} ^{{x}} \:\:\:\frac{{arctan}\left({u}\:{tan}\left(\frac{{t}}{\mathrm{2}}\right)\right)}{{u}}\:{du}\:+{c} \\ $$$${I}\left({t}\right)\:=\mathrm{2}\:{f}\left(\mathrm{1}\right)\:=\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{{arctan}\left({utan}\left(\frac{{t}}{\mathrm{2}}\right)\right)}{{u}}{du}\:+{c}\:\:…{be}\:{continued}… \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 07/Oct/18
![∫((x(1+tan^2 (x/2)))/(2tan(x/2)))dx (1/2)∫x(((sec^2 (x/2))/(tan(x/2))))dx (1/2)x∫((sec^2 (x/2))/(tan(x/2)))dx−∫[((d((x/2)))/dx)∫((sec^2 (x/2))/(tan(x/2)))dx]dx xln∣tan(x/2)∣−∫ln∣tan(x/2)∣dx ....pls check the question...](https://www.tinkutara.com/question/Q45012.png)
$$\int\frac{{x}\left(\mathrm{1}+{tan}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}\right)}{\mathrm{2}{tan}\frac{{x}}{\mathrm{2}}}{dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int{x}\left(\frac{{sec}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}{{tan}\frac{{x}}{\mathrm{2}}}\right){dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{x}\int\frac{{sec}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}{{tan}\frac{{x}}{\mathrm{2}}}{dx}−\int\left[\frac{{d}\left(\frac{{x}}{\mathrm{2}}\right)}{{dx}}\int\frac{{sec}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}{{tan}\frac{{x}}{\mathrm{2}}}{dx}\right]{dx} \\ $$$${xln}\mid{tan}\frac{{x}}{\mathrm{2}}\mid−\int{ln}\mid{tan}\frac{{x}}{\mathrm{2}}\mid{dx} \\ $$$$….{pls}\:{check}\:{the}\:{question}… \\ $$$$ \\ $$$$ \\ $$$$ \\ $$