Misc 22 - Chapter 7 Class 12 Integrals

Transcript

Misc 22 Integrate the function tan^(−1)⁡√((1 − 𝑥)/(1 + 𝑥)) Let x = cos 2𝜃 𝑑𝑥/𝑑𝜃=−2 sin⁡〖2𝜃 〗 dx = −2 sin 2𝜃 d𝜃 Substituting, ∫1▒〖tan^(−1)⁡√((1 − 𝑥)/(1 + 𝑥)) 𝑑𝑥〗 = ∫1▒〖𝑡𝑎𝑛〗^(−1) √((1 − cos⁡2𝜃)/(1 + cos⁡2𝜃 ))×(−2 sin⁡〖2 𝜃)〗 𝑑 𝜃 = −2∫1▒〖𝑡𝑎𝑛〗^(−1) √((1 − (1 − 2〖𝑠𝑖𝑛〗^2 𝜃))/(1 + (2〖𝑐𝑜𝑠〗^2 𝜃 − 1) ))×("sin 2𝜃 d𝜃 " ) = −2∫1▒〖𝑡𝑎𝑛〗^(−1) √((sin^2⁡𝜃/cos^2⁡𝜃 ) )×sin⁡〖2𝜃 𝑑𝜃〗 = −2∫1▒〖𝑡𝑎𝑛〗^(−1) (sin⁡𝜃/cos⁡𝜃 )×sin⁡〖2𝜃 𝑑𝜃〗 = −2∫1▒〖𝑡𝑎𝑛〗^(−1) (𝑡𝑎𝑛⁡𝜃 )×sin⁡〖2𝜃 𝑑𝜃〗 = − 2 ∫1▒𝜃 sin⁡〖2𝜃 𝑑𝜃〗 =−2(𝜃∫1▒〖sin⁡2𝜃 𝑑𝜃−(∫1▒𝑑(𝜃)/𝑑𝜃 ∫1▒sin⁡2𝜃 𝑑𝜃) 𝑑𝜃〗) =−2(𝜃((−cos⁡2𝜃)/2)−∫1▒1((−cos⁡2𝜃)/2) 𝑑𝜃) =−2(−𝜃(cos⁡2𝜃/2)⤶+∫1▒cos⁡2𝜃/2 𝑑𝜃) =−2(−(𝜃 cos⁡2𝜃)/2⤶+sin⁡2𝜃/4) 1/2 〖𝑐𝑜𝑠〗^(−1) (𝑥)=𝜃 𝜃 = 1/2 〖𝑐𝑜𝑠〗^(−1) 𝑥 𝑥^2=〖𝑐𝑜𝑠〗^2 2𝜃 𝑥^2=1−〖𝑠𝑖𝑛〗^2 2𝜃 〖𝑠𝑖𝑛〗^2 2𝜃="1 − " 𝑥^2 sin 2𝜃 = √(1−𝑥^2 ) Now, x = cos 2𝜃 Putting the values = −2 (−1/2 (1/2 〖𝑐𝑜𝑠〗^(−1) 𝑥)𝑥+√(1 − 𝑥^2 )/4) = −2 (√(1 − 𝑥^2 )/4−(𝑥 〖𝑐𝑜𝑠〗^(−1) 𝑥)/4)+ C = 𝟏/𝟐 (𝒙〖 𝒄𝒐𝒔〗^(−𝟏) 𝒙−√(𝟏−𝒙^𝟐 ) )+ C