Question Number 155686 by john_santu last updated on 03/Oct/21
$$\:\:\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \frac{{dx}}{\mathrm{1}+\mathrm{tan}\:{x}}\:=? \\ $$
Answered by peter frank last updated on 03/Oct/21
$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{cos}\:\mathrm{x}}{\mathrm{cos}\:\mathrm{x}+\mathrm{sin}\:\mathrm{x}}\mathrm{dx} \\ $$$$\mathrm{by}\:\mathrm{t}−\mathrm{substitution} \\ $$$$\mathrm{cos}\:\mathrm{x}=\frac{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }{\mathrm{1}+\mathrm{t}^{\mathrm{2}} } \\ $$$$\mathrm{sin}\:\mathrm{x}=\frac{\mathrm{2t}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} } \\ $$$$\mathrm{t}=\mathrm{tan}\:\frac{\mathrm{x}}{\mathrm{2}} \\ $$
Answered by puissant last updated on 03/Oct/21
$${Q}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{dx}}{\mathrm{1}+{tan}\:{x}}\:\:;\:{x}=\frac{\pi}{\mathrm{2}}−{u}\:\rightarrow\:{dx}=−{du} \\ $$$$\Rightarrow\:{Q}=\int_{\frac{\pi}{\mathrm{2}}} ^{\mathrm{0}} \frac{\left(−{du}\right)}{\mathrm{1}+{tan}\left(\frac{\pi}{\mathrm{2}}−{u}\right)}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{du}}{\mathrm{1}+\frac{\mathrm{1}}{{tanu}}} \\ $$$$=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{tan}\:{u}}{\mathrm{1}+{tan}\:{u}}{du}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{1}+{tan}\:{u}\:−\mathrm{1}}{\mathrm{1}+{tan}\:{u}}{du} \\ $$$$\Rightarrow\:\mathrm{2}{Q}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {du}\:\Rightarrow\:{Q}=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\pi}{\mathrm{2}}=\frac{\pi}{\mathrm{4}}.. \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\therefore\because\:\:\:{Q}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{dx}}{\mathrm{1}+{tan}\:{x}}\:=\:\frac{\pi}{\mathrm{4}}… \\ $$