find-the-value-of-0-pi-2-dx-1-tanx-2-

Question Number 63089 by mathmax by abdo last updated on 28/Jun/19

find the value of ∫_0 ^(π/2) (dx/(1+(tanx)^(√2) )) .

$${find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{dx}}{\mathrm{1}+\left({tanx}\right)^{\sqrt{\mathrm{2}}} }\:. \\ $$

Commented by edafe ovwie last updated on 29/Jun/19

using ∫_b ^a f(x)dx=∫_b ^a f(a+b−x)dx let 𝛀=∫_0 ^(π/2) (1/(1+(tanx)^(√2) ))dx.....1 𝛀=∫_0 ^(π/2) (1/(1+(tan((𝛑/2)−x))^(√2) ))dx tan((π/2)−x)=cotx=(1/(tanx)) 𝛀=∫_0 ^(π/2) (1/(1+(cotx)^(√2) ))dx 𝛀=∫_0 ^(π/2) (((tanx)^(√x) )/(1+(tanx)^(√2) ))dx...2 add 1 and 2 2𝛀=∫_0 ^(π/2) 1dx 2Ω=(π/2) Ω=(π/4) ∫_0 ^(π/2) (1/(1+(tanx)^(√2) ))dx=(π/4)

$$\boldsymbol{{using}} \\ $$$$\int_{\boldsymbol{{b}}} ^{\boldsymbol{{a}}} \boldsymbol{{f}}\left(\boldsymbol{{x}}\right)\boldsymbol{{dx}}=\int_{\boldsymbol{{b}}} ^{\boldsymbol{{a}}} \boldsymbol{{f}}\left(\boldsymbol{{a}}+\boldsymbol{{b}}−\boldsymbol{{x}}\right)\boldsymbol{{dx}} \\ $$$${let} \\ $$$$\boldsymbol{\Omega}=\overset{\frac{\pi}{\mathrm{2}}} {\int}_{\mathrm{0}} \frac{\mathrm{1}}{\mathrm{1}+\left({tanx}\overset{\sqrt{\mathrm{2}}} {\right)}}{dx}…..\mathrm{1} \\ $$$$\boldsymbol{\Omega}=\overset{\frac{\pi}{\mathrm{2}}} {\int}_{\mathrm{0}} \frac{\mathrm{1}}{\mathrm{1}+\left({tan}\left(\frac{\boldsymbol{\pi}}{\mathrm{2}}−{x}\right)\overset{\sqrt{\mathrm{2}}} {\right)}}{dx} \\ $$$${tan}\left(\frac{\pi}{\mathrm{2}}−{x}\right)={cotx}=\frac{\mathrm{1}}{{tanx}} \\ $$$$\boldsymbol{\Omega}=\overset{\frac{\pi}{\mathrm{2}}} {\int}_{\mathrm{0}} \frac{\mathrm{1}}{\mathrm{1}+\left({cotx}\overset{\sqrt{\mathrm{2}}} {\right)}}{dx} \\ $$$$\boldsymbol{\Omega}=\overset{\frac{\pi}{\mathrm{2}}} {\int}_{\mathrm{0}} \frac{\left({tanx}\right)^{\sqrt{{x}}} }{\mathrm{1}+\left({tanx}\overset{\sqrt{\mathrm{2}}} {\right)}}{dx}…\mathrm{2} \\ $$$${add}\:\mathrm{1}\:{and}\:\mathrm{2} \\ $$$$\mathrm{2}\boldsymbol{\Omega}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{1}{dx} \\ $$$$\mathrm{2}\Omega=\frac{\pi}{\mathrm{2}} \\ $$$$\Omega=\frac{\pi}{\mathrm{4}} \\ $$$$\overset{\frac{\pi}{\mathrm{2}}} {\int}_{\mathrm{0}} \frac{\mathrm{1}}{\mathrm{1}+\left({tanx}\overset{\sqrt{\mathrm{2}}} {\right)}}{dx}=\frac{\pi}{\mathrm{4}} \\ $$

Commented by mathmax by abdo last updated on 29/Jun/19

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

Commented by mathmax by abdo last updated on 03/Jul/19

let A =∫_0 ^(π/2) (dx/(1+(tanx)^(√2) )) changement x =(π/2)−t give A =−∫_0 ^(π/2) ((−dt)/(1+ ((1/(tant)))^(√2) )) = ∫_0 ^(π/2) (((tant)^(√2) )/(1+(tant)^(√2) )) dt =∫_0 ^(π/2) ((1+(tant)^(√2) −1)/(1+(tant)^(√2) ))dt =(π/2) −A ⇒ 2A =(π/2) ⇒ A =(π/4) .

$${let}\:{A}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{dx}}{\mathrm{1}+\left({tanx}\right)^{\sqrt{\mathrm{2}}} }\:\:{changement}\:{x}\:=\frac{\pi}{\mathrm{2}}−{t}\:{give} \\ $$$${A}\:=−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{−{dt}}{\mathrm{1}+\:\left(\frac{\mathrm{1}}{{tant}}\right)^{\sqrt{\mathrm{2}}} }\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\frac{\left({tant}\right)^{\sqrt{\mathrm{2}}} }{\mathrm{1}+\left({tant}\right)^{\sqrt{\mathrm{2}}} }\:{dt}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{1}+\left({tant}\right)^{\sqrt{\mathrm{2}}} −\mathrm{1}}{\mathrm{1}+\left({tant}\right)^{\sqrt{\mathrm{2}}} }{dt} \\ $$$$=\frac{\pi}{\mathrm{2}}\:−{A}\:\Rightarrow\:\mathrm{2}{A}\:=\frac{\pi}{\mathrm{2}}\:\Rightarrow\:{A}\:=\frac{\pi}{\mathrm{4}}\:. \\ $$

Facebook Tweet Pin

find-the-value-of-0-pi-2-dx-1-tanx-2-

Leave a Reply Cancel reply