Menu Close

Question-201515




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)
$$\boldsymbol{{Solution}}:\:\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \:\frac{\mathrm{1}}{\mathrm{1}+\left(\frac{\mathrm{1}}{\boldsymbol{{tanx}}}\right)^{\sqrt{\mathrm{2}}} }\boldsymbol{{dx}}\:\:\Leftrightarrow\:\:\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \:\frac{\mathrm{1}}{\frac{\left(\boldsymbol{{tanx}}\right)^{\sqrt{\mathrm{2}}} +\mathrm{1}}{\left(\boldsymbol{{tanx}}\right)^{\sqrt{\mathrm{2}}} }}\boldsymbol{{dx}} \\ $$$$\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \:\frac{\left(\boldsymbol{{tanx}}\right)^{\sqrt{\mathrm{2}}} }{\mathrm{1}+\left(\boldsymbol{{tanx}}\right)^{\sqrt{\mathrm{2}}} }\boldsymbol{{dx}}\:\:\:\:\boldsymbol{{let}}\:\boldsymbol{{y}}=\frac{\boldsymbol{\pi}}{\mathrm{2}}−\boldsymbol{{x}}\:\:\:\boldsymbol{{dy}}=−\boldsymbol{{dx}} \\ $$$$\boldsymbol{{when}}\:\boldsymbol{{x}}=\frac{\boldsymbol{\pi}}{\mathrm{2}\:}\:\:\:\boldsymbol{{y}}=\mathrm{0}\:\:\boldsymbol{{and}}\:\:\boldsymbol{{when}}\:\boldsymbol{{x}}=\mathrm{0}\:\:\boldsymbol{{y}}=\frac{\boldsymbol{\pi}}{\mathrm{2}} \\ $$$$\boldsymbol{{I}}=\int_{\frac{\boldsymbol{\pi}}{\mathrm{2}}} ^{\mathrm{0}} \:\frac{\left[\boldsymbol{{tan}}\left(\frac{\boldsymbol{\pi}}{\mathrm{2}}−\boldsymbol{{y}}\right)\right]^{\sqrt{\mathrm{2}}} }{\mathrm{1}+\left[\boldsymbol{{tan}}\left(\frac{\boldsymbol{\pi}}{\mathrm{2}}−\boldsymbol{{y}}\right)\right]^{\sqrt{\mathrm{2}}} }\left(−\boldsymbol{{dy}}\right)\:\:\:\:\Leftrightarrow\:\:\:\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \:\frac{\left(\boldsymbol{{coty}}\right)^{\sqrt{\mathrm{2}}} }{\mathrm{1}+\left(\boldsymbol{{coty}}\right)^{\sqrt{\mathrm{2}}} }\boldsymbol{{dy}} \\ $$$$\boldsymbol{{Nb}}:\:\boldsymbol{{tan}}\left(\frac{\boldsymbol{\pi}}{\mathrm{2}}−\boldsymbol{{y}}\right)=\boldsymbol{{coty}}\:\:\:\boldsymbol{{and}}\:\boldsymbol{{changing}}\:\boldsymbol{{of}}\:\boldsymbol{{variable}} \\ $$$$\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \:\frac{\left(\boldsymbol{{cotx}}\right)^{\sqrt{\mathrm{2}}} }{\mathrm{1}+\left(\boldsymbol{{cotx}}\right)^{\sqrt{\mathrm{2}}} }\boldsymbol{{dx}}\:\:\:\:\:\left(\boldsymbol{{add}}\:\boldsymbol{{the}}\:\boldsymbol{{two}}\:\boldsymbol{{integral}}\right) \\ $$$$\boldsymbol{{I}}+\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \:\frac{\mathrm{1}}{\mathrm{1}+\left(\boldsymbol{{cotx}}\right)^{\sqrt{\mathrm{2}}} }\boldsymbol{{dx}}\:+\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \:\frac{\left(\boldsymbol{{cotx}}\right)^{\sqrt{\mathrm{2}}} }{\mathrm{1}+\left(\boldsymbol{{cotx}}\right)^{\sqrt{\mathrm{2}}} }\boldsymbol{{dx}} \\ $$$$\mathrm{2}\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \:\frac{\mathrm{1}+\left(\boldsymbol{{cotx}}\right)^{\sqrt{\mathrm{2}}} }{\mathrm{1}+\left(\boldsymbol{{cotx}}\right)^{\sqrt{\mathrm{2}}} }\boldsymbol{{dx}}\:\:\Leftrightarrow\:\:\mathrm{2}\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \mathrm{1}\boldsymbol{{dx}} \\ $$$$\mathrm{2}\boldsymbol{{I}}=\boldsymbol{{x}}\mid_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} +\boldsymbol{{C}} \\ $$$$\mathrm{2}\boldsymbol{{I}}=\left(\frac{\boldsymbol{\pi}}{\mathrm{2}}−\mathrm{0}\right) \\ $$$$\boldsymbol{{I}}=\frac{\boldsymbol{\pi}}{\mathrm{4}} \\ $$$$ \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *