Integration Questions

Question Number 55930 by ajfour last updated on 06/Mar/19

$$\int_{\mathrm{0}} ^{\:\:\pi} \frac{{x}\mathrm{tan}\:{x}}{\mathrm{sec}\:{x}+\mathrm{tan}\:{x}}{dx}\:=\:\left({is}\:{it}\:\frac{\pi^{\mathrm{2}} }{\mathrm{2}}−\pi\right)? \\$$

Answered by tanmay.chaudhury50@gmail.com last updated on 06/Mar/19

$$\int_{\mathrm{0}} ^{\pi} \frac{{xsinx}}{\mathrm{1}+{sinx}}{dx} \\$$$${I}=\int_{\mathrm{0}} ^{\pi} \frac{\left(\pi−{x}\right){sin}\left(\pi−{x}\right)}{\mathrm{1}+{sin}\left(\pi−{x}\right)}{dx} \\$$$$=\int_{\mathrm{0}} ^{\pi} \frac{\left(\pi−{x}\right){sinx}}{\mathrm{1}+{sinx}}{dx} \\$$$$\mathrm{2}{I}=\int_{\mathrm{0}} ^{\pi} \frac{\pi{sinx}−{xsinx}+{xsinx}}{\mathrm{1}+{sinx}}{dx} \\$$$$\mathrm{2}{I}=\pi\int_{\mathrm{0}} ^{\pi} \frac{{sinx}}{\mathrm{1}+{sinx}}{dx} \\$$$$=\pi\int_{\mathrm{0}} ^{\pi} \left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}+{sinx}}\right){dx} \\$$$$\frac{\mathrm{2}{I}}{\pi}=\int_{\mathrm{0}} ^{\pi} {dx}−\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{1}−{sinx}}{{cos}^{\mathrm{2}} {x}}{dx} \\$$$$\frac{\mathrm{2}{I}}{\pi}=\int_{\mathrm{0}} ^{\pi} {dx}−\int_{\mathrm{0}} ^{\pi} {sec}^{\mathrm{2}} {xdx}+\int_{\mathrm{0}} ^{\pi} {secxtanxdx} \\$$$$\frac{\mathrm{2}{I}}{\pi}=\mid{x}−{tanx}+{secx}\mid_{\mathrm{0}} ^{\pi} \\$$$$\frac{\mathrm{2}{I}}{\pi}=\left(\pi−\mathrm{0}−\mathrm{1}\right)−\left(\mathrm{0}−\mathrm{0}+\mathrm{1}\right) \\$$$$\frac{\mathrm{2}{I}}{\pi}=\pi−\mathrm{1}−\mathrm{1} \\$$$$\mathrm{2}{I}=\pi^{\mathrm{2}} −\mathrm{2}\pi \\$$$${I}=\frac{\pi^{\mathrm{2}} }{\mathrm{2}}−\pi \\$$

Commented by ajfour last updated on 06/Mar/19

$${Thanks}\:{Sir},\:{great}! \\$$

Commented by tanmay.chaudhury50@gmail.com last updated on 06/Mar/19

$${thank}\:{you}\:{sir}... \\$$

Commented by maxmathsup by imad last updated on 09/Mar/19

$${I}\:=\int_{\mathrm{0}} ^{\pi} \:\frac{{x}\left(\mathrm{1}+{sinx}−\mathrm{1}\right)}{\mathrm{1}+{sinx}}{dx}\:=\int_{\mathrm{0}} ^{\pi} \:{xdx}\:−\int_{\mathrm{0}} ^{\pi} \:\:\frac{{x}}{\mathrm{1}+{sinx}}\:{dx}\:{but} \\$$$$\int_{\mathrm{0}} ^{\pi} \:{xdx}\:=\left[\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right]_{\mathrm{0}} ^{\pi} =\frac{\pi^{\mathrm{2}} }{\mathrm{2}}\:\:\:{and}\:\int_{\mathrm{0}} ^{\pi} \:\:\frac{{x}}{\mathrm{1}+{sinx}}\:{dx}\:=_{{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}} \:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{\mathrm{2}{arctan}\left({t}\right)}{\mathrm{1}+\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }}\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\$$$$=\mathrm{4}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{arctan}\left({t}\right)}{\mathrm{1}+{t}^{\mathrm{2}} \:+\mathrm{2}{t}}\:{dt}\:=\mathrm{4}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{arctan}\left({t}\right)}{\left({t}+\mathrm{1}\right)^{\mathrm{2}} }\:{dt}\:\:\:\:\:\left(\:{and}\:{by}\:{parts}\right) \\$$$$=\mathrm{4}\left\{\:\left[−\frac{{arctan}\left({t}\right)}{{t}+\mathrm{1}}\right]_{\mathrm{0}} ^{\infty} \:+\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{1}}{\left({t}+\mathrm{1}\right)\left({t}^{\mathrm{2}} +\mathrm{1}\right)}{dt}\right\} \\$$$$=\mathrm{4}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{{dt}}{\left({t}+\mathrm{1}\right)\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)}\:{dt}\:\:{let}\:{decompose}\:{F}\left({t}\right)\:=\frac{\mathrm{1}}{\left({t}+\mathrm{1}\right)\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)}\:\Rightarrow \\$$$${F}\left({t}\right)\:=\frac{{a}}{{t}+\mathrm{1}}\:+\frac{{bt}\:+{c}}{{t}^{\mathrm{2}} \:+\mathrm{1}} \\$$$${a}\:={lim}_{{t}\rightarrow−\mathrm{1}} \left({t}+\mathrm{1}\right){F}\left({t}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\$$$${lim}_{{t}\rightarrow+\infty} {tF}\left({t}\right)\:=\:\mathrm{0}\:={a}+{b}\:\Rightarrow{b}\:=−{a}\:=−\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow{F}\left({t}\right)=\frac{\mathrm{1}}{\mathrm{2}\left({t}+\mathrm{1}\right)}\:+\frac{−\frac{{t}}{\mathrm{2}}+{c}}{{t}^{\mathrm{2}} \:+\mathrm{1}} \\$$$${F}\left(\mathrm{0}\right)\:=\mathrm{1}\:=\frac{\mathrm{1}}{\mathrm{2}}\:+{c}\:\Rightarrow{c}=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow{F}\left({t}\right)=\frac{\mathrm{1}}{\mathrm{2}\left({t}+\mathrm{1}\right)}\:−\frac{\mathrm{1}}{\mathrm{2}}\:\frac{{t}−\mathrm{1}}{{t}^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow \\$$$$\int_{\mathrm{0}} ^{\infty} \:{F}\left({t}\right){dt}\:=\frac{\mathrm{1}}{\mathrm{2}}\left[{ln}\mid{t}+\mathrm{1}\mid−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)\right]_{\mathrm{0}} ^{+\infty} \:+\frac{\mathrm{1}}{\mathrm{2}}\left[{arctan}\left({t}\right)\right]_{\mathrm{0}} ^{+\infty} \\$$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[{ln}\mid\frac{{t}+\mathrm{1}}{\sqrt{{t}^{\mathrm{2}} \:+\mathrm{1}}}\mid\right]_{\mathrm{0}} ^{\infty} \:+\frac{\pi}{\mathrm{4}}\:=\frac{\pi}{\mathrm{4}}\:\Rightarrow \\$$$${I}\:=\frac{\pi^{\mathrm{2}} }{\mathrm{2}}\:−\mathrm{4}\left(\frac{\pi}{\mathrm{4}}\right)\:=\frac{\pi^{\mathrm{2}} }{\mathrm{2}}−\pi\:\:. \\$$$$\\$$$$\\$$

Answered by MJS last updated on 06/Mar/19

$$\int\frac{{x}\mathrm{tan}\:{x}}{\mathrm{sec}\:{x}\:+\mathrm{tan}\:{x}}{dx}=\int\frac{{x}\mathrm{sin}\:{x}}{\mathrm{1}+\mathrm{sin}\:{x}}{dx}=\int{xdx}−\int\frac{{x}}{\mathrm{1}+\mathrm{sin}\:{x}}{dx}= \\$$$$=\frac{{x}^{\mathrm{2}} }{\mathrm{2}}−\int\frac{{x}}{\mathrm{1}+\mathrm{sin}\:{x}}{dx}= \\$$$$\:\:\:\:\:\int\frac{{x}}{\mathrm{1}+\mathrm{sin}\:{x}}{dx}= \\$$$$\:\:\:\:\:\:\:\:\:\:\left[\int{u}'{v}={uv}−\int{uv}';\:{u}'=\frac{\mathrm{1}}{\mathrm{1}+\mathrm{sin}\:{x}};\:{v}={x}\right. \\$$$$\left.\:\:\:\:\:\:\:\:\:\:\:{u}=−\frac{\mathrm{cos}\:{x}}{\mathrm{1}+\mathrm{sin}\:{x}};\:{v}'=\mathrm{1}\right] \\$$$$\:\:\:\:\:=−\frac{{x}\mathrm{cos}\:{x}}{\mathrm{1}+\mathrm{sin}\:{x}}+\int\frac{\mathrm{cos}\:{x}}{\mathrm{1}+\mathrm{sin}\:{x}}{dx}= \\$$$$\:\:\:\:\:=−\frac{{x}\mathrm{cos}\:{x}}{\mathrm{1}+\mathrm{sin}\:{x}}+\mathrm{ln}\:\left(\mathrm{1}+\mathrm{sin}\:{x}\right) \\$$$$=\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{{x}\mathrm{cos}\:{x}}{\mathrm{1}+\mathrm{sin}\:{x}}−\mathrm{ln}\:\left(\mathrm{1}+\mathrm{sin}\:{x}\right)\:+{C} \\$$$$\Rightarrow\:\mathrm{your}\:\mathrm{value}\:\mathrm{is}\:\mathrm{true} \\$$

Commented by ajfour last updated on 06/Mar/19

$${Thanks}\:{MJS}\:{Sir}! \\$$