Question Number 201516 by sonukgindia last updated on 08/Dec/23
Answered by Calculusboy last updated on 08/Dec/23
![Solution: let y=(𝛑/2)−x dy=−dx when x=(𝛑/2) y=0 and when x=0 y=(𝛑/2) I=∫_(𝛑/2) ^0 (1/(1+[tan((𝛑/2)−y)]^n ))(−dy) ⇔ I=∫_0 ^(𝛑/2) (1/(1+(coty)))dy Nb: tan((𝛑/2)−y)=coty changing of variable and (cotx=(1/(tanx))) I=∫_0 ^(𝛑/2) (1/(1+((1/(tanx)))^n )) dx ⇔ I=∫_0 ^(𝛑/2) (1/(((tanx)^n +1)/((tanx)^n )))dx I=∫_0 ^(𝛑/2) (((tanx)^n )/(1+(tanx)^n ))dx (add the two integral) I+I=∫_0 ^(𝛑/2) (1/(1+(tanx)^n ))dx+∫_0 ^(𝛑/2) (((tanx)^n )/(1+(tanx)^n ))dx 2I=∫_0 ^(𝛑/2) ((1+(tanx)^n )/(1+(tanx)^n ))dx ⇔ 2I=∫_0 ^(𝛑/2) 1dx 2I=x∣_0 ^(𝛑/2) +C 2I=((𝛑/2)−0) I=(𝛑/4)](https://www.tinkutara.com/question/Q201518.png)
$$\boldsymbol{{Solution}}:\:\boldsymbol{{let}}\:\boldsymbol{{y}}=\frac{\boldsymbol{\pi}}{\mathrm{2}}−\boldsymbol{{x}}\:\:\:\boldsymbol{{dy}}=−\boldsymbol{{dx}} \\ $$$${when}\:\boldsymbol{{x}}=\frac{\boldsymbol{\pi}}{\mathrm{2}}\:\:\:\boldsymbol{{y}}=\mathrm{0}\:\:\boldsymbol{{and}}\:\boldsymbol{{when}}\:\boldsymbol{{x}}=\mathrm{0}\:\:\boldsymbol{{y}}=\frac{\boldsymbol{\pi}}{\mathrm{2}} \\ $$$$\boldsymbol{{I}}=\int_{\frac{\boldsymbol{\pi}}{\mathrm{2}}} ^{\mathrm{0}} \:\frac{\mathrm{1}}{\mathrm{1}+\left[\boldsymbol{{tan}}\left(\frac{\boldsymbol{\pi}}{\mathrm{2}}−\boldsymbol{{y}}\right)\right]^{\boldsymbol{{n}}} }\left(−\boldsymbol{{dy}}\right)\:\:\:\Leftrightarrow\:\:\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \:\frac{\mathrm{1}}{\mathrm{1}+\left(\boldsymbol{{coty}}\right)}\boldsymbol{{dy}}\:\:\:\:\boldsymbol{{Nb}}:\:\boldsymbol{{tan}}\left(\frac{\boldsymbol{\pi}}{\mathrm{2}}−\boldsymbol{{y}}\right)=\boldsymbol{{coty}} \\ $$$$\boldsymbol{{changing}}\:\boldsymbol{{of}}\:\boldsymbol{{variable}}\:\:\boldsymbol{{and}}\:\left(\boldsymbol{{cotx}}=\frac{\mathrm{1}}{\boldsymbol{{tanx}}}\right) \\ $$$$\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \:\frac{\mathrm{1}}{\mathrm{1}+\left(\frac{\mathrm{1}}{\boldsymbol{{tanx}}}\right)^{\boldsymbol{{n}}} }\:\boldsymbol{{dx}}\:\:\:\Leftrightarrow\:\:\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \:\frac{\mathrm{1}}{\frac{\left(\boldsymbol{{tanx}}\right)^{\boldsymbol{{n}}} +\mathrm{1}}{\left(\boldsymbol{{tanx}}\right)^{\boldsymbol{{n}}} }}\boldsymbol{{dx}} \\ $$$$\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \:\frac{\left(\boldsymbol{{tanx}}\right)^{\boldsymbol{{n}}} }{\mathrm{1}+\left(\boldsymbol{{tanx}}\right)^{\boldsymbol{{n}}} }\boldsymbol{{dx}}\:\:\left(\boldsymbol{{add}}\:\boldsymbol{{the}}\:\boldsymbol{{two}}\:\boldsymbol{{integral}}\right) \\ $$$$\boldsymbol{{I}}+\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \frac{\mathrm{1}}{\mathrm{1}+\left(\boldsymbol{{tanx}}\right)^{\boldsymbol{{n}}} }\boldsymbol{{dx}}+\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \frac{\left(\boldsymbol{{tanx}}\right)^{\boldsymbol{{n}}} }{\mathrm{1}+\left(\boldsymbol{{tanx}}\right)^{\boldsymbol{{n}}} }\boldsymbol{{dx}} \\ $$$$\mathrm{2}\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \:\frac{\mathrm{1}+\left(\boldsymbol{{tanx}}\right)^{\boldsymbol{{n}}} }{\mathrm{1}+\left(\boldsymbol{{tanx}}\right)^{\boldsymbol{{n}}} }\boldsymbol{{dx}}\:\:\Leftrightarrow\:\:\mathrm{2}\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \mathrm{1}\boldsymbol{{dx}} \\ $$$$\mathrm{2}\boldsymbol{{I}}=\boldsymbol{{x}}\mid_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} +\boldsymbol{{C}} \\ $$$$\mathrm{2}\boldsymbol{{I}}=\left(\frac{\boldsymbol{\pi}}{\mathrm{2}}−\mathrm{0}\right) \\ $$$$\boldsymbol{{I}}=\frac{\boldsymbol{\pi}}{\mathrm{4}} \\ $$$$ \\ $$