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0-pi-2-dx-1-tan-x-




Question Number 155686 by john_santu last updated on 03/Oct/21
  ∫_0 ^(π/2) (dx/(1+tan x)) =?
$$\:\:\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \frac{{dx}}{\mathrm{1}+\mathrm{tan}\:{x}}\:=? \\ $$
Answered by peter frank last updated on 03/Oct/21
∫_0 ^(π/2) ((cos x)/(cos x+sin x))dx  by t−substitution  cos x=((1−t^2 )/(1+t^2 ))  sin x=((2t)/(1+t^2 ))  t=tan (x/2)
$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{cos}\:\mathrm{x}}{\mathrm{cos}\:\mathrm{x}+\mathrm{sin}\:\mathrm{x}}\mathrm{dx} \\ $$$$\mathrm{by}\:\mathrm{t}−\mathrm{substitution} \\ $$$$\mathrm{cos}\:\mathrm{x}=\frac{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }{\mathrm{1}+\mathrm{t}^{\mathrm{2}} } \\ $$$$\mathrm{sin}\:\mathrm{x}=\frac{\mathrm{2t}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} } \\ $$$$\mathrm{t}=\mathrm{tan}\:\frac{\mathrm{x}}{\mathrm{2}} \\ $$
Answered by puissant last updated on 03/Oct/21
Q=∫_0 ^(π/2) (dx/(1+tan x))  ; x=(π/2)−u → dx=−du  ⇒ Q=∫_(π/2) ^0 (((−du))/(1+tan((π/2)−u)))=∫_0 ^(π/2) (du/(1+(1/(tanu))))  = ∫_0 ^(π/2) ((tan u)/(1+tan u))du = ∫_0 ^(π/2) ((1+tan u −1)/(1+tan u))du  ⇒ 2Q=∫_0 ^(π/2) du ⇒ Q=(1/2)×(π/2)=(π/4)..            ∴∵   Q = ∫_0 ^(π/2) (dx/(1+tan x)) = (π/4)...
$${Q}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{dx}}{\mathrm{1}+{tan}\:{x}}\:\:;\:{x}=\frac{\pi}{\mathrm{2}}−{u}\:\rightarrow\:{dx}=−{du} \\ $$$$\Rightarrow\:{Q}=\int_{\frac{\pi}{\mathrm{2}}} ^{\mathrm{0}} \frac{\left(−{du}\right)}{\mathrm{1}+{tan}\left(\frac{\pi}{\mathrm{2}}−{u}\right)}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{du}}{\mathrm{1}+\frac{\mathrm{1}}{{tanu}}} \\ $$$$=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{tan}\:{u}}{\mathrm{1}+{tan}\:{u}}{du}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{1}+{tan}\:{u}\:−\mathrm{1}}{\mathrm{1}+{tan}\:{u}}{du} \\ $$$$\Rightarrow\:\mathrm{2}{Q}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {du}\:\Rightarrow\:{Q}=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\pi}{\mathrm{2}}=\frac{\pi}{\mathrm{4}}.. \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\therefore\because\:\:\:{Q}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{dx}}{\mathrm{1}+{tan}\:{x}}\:=\:\frac{\pi}{\mathrm{4}}… \\ $$

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