Question Number 201517 by sonukgindia last updated on 08/Dec/23
Answered by Calculusboy last updated on 08/Dec/23
![Solution: let 𝛉=(𝛑/2)−x d𝛉=−dx when x=(𝛑/2) 𝛉=0 and when x=0 𝛉=(𝛑/2) 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 I=∫_(𝛑/2) ^0 (([tan((𝛑/2)−𝛉)]^n )/(1+[tan((𝛑/2)−𝛉)]^n ))(−d𝛉) Nb: tan((𝛑/2)−𝛉)=cot𝛉 I=∫_0 ^(𝛑/2) (((cot𝛉)^n )/(1+(cot𝛉)^n ))d𝛉 (changing of variable) I=∫_0 ^(𝛑/2) (((cotx)^n )/(1+(cotx)^n ))dx (add the two integral) I+I=∫_0 ^(𝛑/2) (1/(1+(cotx)^n ))dx+∫_0 ^(𝛑/2) (((cotx)^n )/(1+(cotx)^n ))dx 2I=∫_0 ^(𝛑/2) ((1+(cotx)^n )/(1+(cotx)^n ))dx ⇔ 2I=∫_0 ^(𝛑/2) 1 dx 2I=x∣_0 ^(𝛑/2) +C 2I=((𝛑/2)−0) I=(𝛑/4)](https://www.tinkutara.com/question/Q201520.png)
$$\boldsymbol{{Solution}}:\:\boldsymbol{{let}}\:\boldsymbol{\theta}=\frac{\boldsymbol{\pi}}{\mathrm{2}}−\boldsymbol{{x}}\:\:\boldsymbol{{d}\theta}=−\boldsymbol{{dx}} \\ $$$$\boldsymbol{{when}}\:\boldsymbol{{x}}=\frac{\boldsymbol{\pi}}{\mathrm{2}}\:\:\:\boldsymbol{\theta}=\mathrm{0}\:\:\boldsymbol{{and}}\:\boldsymbol{{when}}\:\boldsymbol{{x}}=\mathrm{0}\:\:\boldsymbol{\theta}=\frac{\boldsymbol{\pi}}{\mathrm{2}} \\ $$$$\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \frac{\mathrm{1}}{\mathrm{1}+\left(\frac{\mathrm{1}}{\boldsymbol{{tanx}}}\right)^{\boldsymbol{{n}}} }\boldsymbol{{dx}}\:\:\Leftrightarrow\:\:\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \frac{\mathrm{1}}{\frac{\left(\boldsymbol{{tanx}}\right)^{\boldsymbol{{n}}} +\mathrm{1}}{\left(\boldsymbol{{tanx}}\right)^{\boldsymbol{{n}}} }}\boldsymbol{{dx}}\:\: \\ $$$$\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \:\frac{\left(\boldsymbol{{tanx}}\right)^{\boldsymbol{{n}}} }{\mathrm{1}+\left(\boldsymbol{{tanx}}\right)^{\boldsymbol{{n}}} }\boldsymbol{{dx}}\:\:\boldsymbol{{I}}=\int_{\frac{\boldsymbol{\pi}}{\mathrm{2}}} ^{\mathrm{0}} \:\frac{\left[\boldsymbol{{tan}}\left(\frac{\boldsymbol{\pi}}{\mathrm{2}}−\boldsymbol{\theta}\right)\right]^{\boldsymbol{{n}}} }{\mathrm{1}+\left[\boldsymbol{{tan}}\left(\frac{\boldsymbol{\pi}}{\mathrm{2}}−\boldsymbol{\theta}\right)\right]^{\boldsymbol{{n}}} }\left(−\boldsymbol{{d}\theta}\right)\:\:\: \\ $$$$\boldsymbol{{Nb}}:\:\boldsymbol{{tan}}\left(\frac{\boldsymbol{\pi}}{\mathrm{2}}−\boldsymbol{\theta}\right)=\boldsymbol{{cot}\theta}\:\:\: \\ $$$$\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \:\frac{\left(\boldsymbol{{cot}\theta}\right)^{\boldsymbol{{n}}} }{\mathrm{1}+\left(\boldsymbol{{cot}\theta}\right)^{\boldsymbol{{n}}} }\boldsymbol{{d}\theta}\:\:\:\left(\boldsymbol{{changing}}\:\boldsymbol{{of}}\:\boldsymbol{{variable}}\right) \\ $$$$\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \:\frac{\left(\boldsymbol{{cotx}}\right)^{\boldsymbol{{n}}} }{\mathrm{1}+\left(\boldsymbol{{cotx}}\right)^{\boldsymbol{{n}}} }\boldsymbol{{dx}}\:\:\:\:\left(\boldsymbol{{add}}\:\boldsymbol{{the}}\:\boldsymbol{{two}}\:\boldsymbol{{integral}}\right) \\ $$$$\boldsymbol{{I}}+{I}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \frac{\mathrm{1}}{\mathrm{1}+\left(\boldsymbol{{cotx}}\right)^{\boldsymbol{{n}}} }\boldsymbol{{dx}}+\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \frac{\left(\boldsymbol{{cotx}}\right)^{\boldsymbol{{n}}} }{\mathrm{1}+\left(\boldsymbol{{cotx}}\right)^{\boldsymbol{{n}}} }\boldsymbol{{dx}} \\ $$$$\mathrm{2}\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \:\frac{\mathrm{1}+\left(\boldsymbol{{cotx}}\right)^{\boldsymbol{{n}}} }{\mathrm{1}+\left(\boldsymbol{{cotx}}\right)^{\boldsymbol{{n}}} }\boldsymbol{{dx}}\:\:\Leftrightarrow\:\:\mathrm{2}\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \mathrm{1}\:\boldsymbol{{dx}}\: \\ $$$$\mathrm{2}\boldsymbol{{I}}=\boldsymbol{{x}}\mid_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \:+\boldsymbol{{C}} \\ $$$$\mathrm{2}\boldsymbol{{I}}=\left(\frac{\boldsymbol{\pi}}{\mathrm{2}}−\mathrm{0}\right) \\ $$$$\boldsymbol{{I}}=\frac{\boldsymbol{\pi}}{\mathrm{4}} \\ $$$$ \\ $$