Question Number 201515 by sonukgindia last updated on 08/Dec/23
Answered by Calculusboy last updated on 08/Dec/23
![Solution: I=∫_0 ^(𝛑/2) (1/(1+((1/(tanx)))^(√2) ))dx ⇔ I=∫_0 ^(𝛑/2) (1/(((tanx)^(√2) +1)/((tanx)^(√2) )))dx I=∫_0 ^(𝛑/2) (((tanx)^(√2) )/(1+(tanx)^(√2) ))dx let y=(𝛑/2)−x dy=−dx when x=(𝛑/(2 )) y=0 and when x=0 y=(𝛑/2) I=∫_(𝛑/2) ^0 (([tan((𝛑/2)−y)]^(√2) )/(1+[tan((𝛑/2)−y)]^(√2) ))(−dy) ⇔ I=∫_0 ^(𝛑/2) (((coty)^(√2) )/(1+(coty)^(√2) ))dy Nb: tan((𝛑/2)−y)=coty and changing of variable I=∫_0 ^(𝛑/2) (((cotx)^(√2) )/(1+(cotx)^(√2) ))dx (add the two integral) I+I=∫_0 ^(𝛑/2) (1/(1+(cotx)^(√2) ))dx +∫_0 ^(𝛑/2) (((cotx)^(√2) )/(1+(cotx)^(√2) ))dx 2I=∫_0 ^(𝛑/2) ((1+(cotx)^(√2) )/(1+(cotx)^(√2) ))dx ⇔ 2I=∫_0 ^(𝛑/2) 1dx 2I=x∣_0 ^(𝛑/2) +C 2I=((𝛑/2)−0) I=(𝛑/4)](https://www.tinkutara.com/question/Q201523.png)
$$\boldsymbol{{Solution}}:\:\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \:\frac{\mathrm{1}}{\mathrm{1}+\left(\frac{\mathrm{1}}{\boldsymbol{{tanx}}}\right)^{\sqrt{\mathrm{2}}} }\boldsymbol{{dx}}\:\:\Leftrightarrow\:\:\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \:\frac{\mathrm{1}}{\frac{\left(\boldsymbol{{tanx}}\right)^{\sqrt{\mathrm{2}}} +\mathrm{1}}{\left(\boldsymbol{{tanx}}\right)^{\sqrt{\mathrm{2}}} }}\boldsymbol{{dx}} \\ $$$$\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \:\frac{\left(\boldsymbol{{tanx}}\right)^{\sqrt{\mathrm{2}}} }{\mathrm{1}+\left(\boldsymbol{{tanx}}\right)^{\sqrt{\mathrm{2}}} }\boldsymbol{{dx}}\:\:\:\:\boldsymbol{{let}}\:\boldsymbol{{y}}=\frac{\boldsymbol{\pi}}{\mathrm{2}}−\boldsymbol{{x}}\:\:\:\boldsymbol{{dy}}=−\boldsymbol{{dx}} \\ $$$$\boldsymbol{{when}}\:\boldsymbol{{x}}=\frac{\boldsymbol{\pi}}{\mathrm{2}\:}\:\:\:\boldsymbol{{y}}=\mathrm{0}\:\:\boldsymbol{{and}}\:\:\boldsymbol{{when}}\:\boldsymbol{{x}}=\mathrm{0}\:\:\boldsymbol{{y}}=\frac{\boldsymbol{\pi}}{\mathrm{2}} \\ $$$$\boldsymbol{{I}}=\int_{\frac{\boldsymbol{\pi}}{\mathrm{2}}} ^{\mathrm{0}} \:\frac{\left[\boldsymbol{{tan}}\left(\frac{\boldsymbol{\pi}}{\mathrm{2}}−\boldsymbol{{y}}\right)\right]^{\sqrt{\mathrm{2}}} }{\mathrm{1}+\left[\boldsymbol{{tan}}\left(\frac{\boldsymbol{\pi}}{\mathrm{2}}−\boldsymbol{{y}}\right)\right]^{\sqrt{\mathrm{2}}} }\left(−\boldsymbol{{dy}}\right)\:\:\:\:\Leftrightarrow\:\:\:\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \:\frac{\left(\boldsymbol{{coty}}\right)^{\sqrt{\mathrm{2}}} }{\mathrm{1}+\left(\boldsymbol{{coty}}\right)^{\sqrt{\mathrm{2}}} }\boldsymbol{{dy}} \\ $$$$\boldsymbol{{Nb}}:\:\boldsymbol{{tan}}\left(\frac{\boldsymbol{\pi}}{\mathrm{2}}−\boldsymbol{{y}}\right)=\boldsymbol{{coty}}\:\:\:\boldsymbol{{and}}\:\boldsymbol{{changing}}\:\boldsymbol{{of}}\:\boldsymbol{{variable}} \\ $$$$\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \:\frac{\left(\boldsymbol{{cotx}}\right)^{\sqrt{\mathrm{2}}} }{\mathrm{1}+\left(\boldsymbol{{cotx}}\right)^{\sqrt{\mathrm{2}}} }\boldsymbol{{dx}}\:\:\:\:\:\left(\boldsymbol{{add}}\:\boldsymbol{{the}}\:\boldsymbol{{two}}\:\boldsymbol{{integral}}\right) \\ $$$$\boldsymbol{{I}}+\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \:\frac{\mathrm{1}}{\mathrm{1}+\left(\boldsymbol{{cotx}}\right)^{\sqrt{\mathrm{2}}} }\boldsymbol{{dx}}\:+\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \:\frac{\left(\boldsymbol{{cotx}}\right)^{\sqrt{\mathrm{2}}} }{\mathrm{1}+\left(\boldsymbol{{cotx}}\right)^{\sqrt{\mathrm{2}}} }\boldsymbol{{dx}} \\ $$$$\mathrm{2}\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \:\frac{\mathrm{1}+\left(\boldsymbol{{cotx}}\right)^{\sqrt{\mathrm{2}}} }{\mathrm{1}+\left(\boldsymbol{{cotx}}\right)^{\sqrt{\mathrm{2}}} }\boldsymbol{{dx}}\:\:\Leftrightarrow\:\:\mathrm{2}\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \mathrm{1}\boldsymbol{{dx}} \\ $$$$\mathrm{2}\boldsymbol{{I}}=\boldsymbol{{x}}\mid_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} +\boldsymbol{{C}} \\ $$$$\mathrm{2}\boldsymbol{{I}}=\left(\frac{\boldsymbol{\pi}}{\mathrm{2}}−\mathrm{0}\right) \\ $$$$\boldsymbol{{I}}=\frac{\boldsymbol{\pi}}{\mathrm{4}} \\ $$$$ \\ $$