Misc 19 - Chapter 7 Class 12 Integrals

Transcript

Misc 19 Integrate the function √((1 − √𝑥)/(1 + √𝑥)) ∫1▒〖√((1 − √𝑥)/(1 + √𝑥)) 𝑑𝑥〗 Let x = 〖𝒄𝒐𝒔〗^𝟐 𝟐𝜽 dx = −4 cos 2𝜃 sin 2𝜃 d𝜃 Substituting, = ∫1▒√((1 − √((〖𝑐𝑜𝑠〗^2 2𝜃) ))/(1 + √((〖𝑐𝑜𝑠〗^2 2𝜃) )))×−4 cos⁡2θ sin⁡2θ 𝑑𝜃 = ∫1▒√((1 − cos⁡2𝜃)/(1 + 𝑐𝑜𝑠 2𝜃))×(−4) cos⁡2θ sin⁡2θ 𝑑𝜃 = −4∫1▒√((1 − (1 − 2〖𝑠𝑖𝑛〗^(2 ) 𝜃))/(1 + (2〖𝑐𝑜𝑠〗^(2 ) 𝜃 − 1) )) cos⁡2θ (2 sin⁡θ cos⁡〖𝜃)〗 𝑑𝜃 = −8∫1▒√((2〖𝑠𝑖𝑛〗^(2 ) 𝜃)/(2〖𝑐𝑜𝑠〗^(2 ) 𝜃)) cos⁡2θ cos⁡𝜃 sin⁡𝜃 𝑑𝜃 = −8∫1▒sin⁡𝜃/cos⁡θ cos⁡2θ cos⁡𝜃 sin⁡𝜃 𝑑𝜃 = −8∫1▒〖〖𝑠𝑖𝑛〗^2 𝜃〗 cos⁡2θ 𝑑𝜃 Using cos⁡2𝑥=1−2〖𝑠𝑖𝑛〗^2 𝑥=2〖𝑐𝑜𝑠〗^2 𝑥−1 sin 2𝑥=2 sin⁡〖𝑥 cos⁡𝑥 〗 = −8∫1▒((1 − cos⁡2θ)/2) cos⁡2θ 𝑑𝜃 = –4 ∫1▒(𝑐𝑜𝑠 2θ−cos^2⁡2θ ) 𝑑𝜃 = 4 ∫1▒(〖𝑐𝑜𝑠〗^2 2θ−cos⁡2θ ) 𝑑𝜃 = 4 ∫1▒〖〖𝑐𝑜𝑠〗^2 2θ〗 𝑑𝜃−4∫1▒cos⁡2θ 𝑑𝜃 = 4 ∫1▒(cos⁡4𝜃 + 1)/2 𝑑𝜃 − 4∫1▒〖𝑐𝑜𝑠 2𝜃〗 𝑑𝜃 = 2 ∫1▒〖(cos⁡4𝜃 + 1)〗 𝑑𝜃 − 4∫1▒〖𝑐𝑜𝑠 2𝜃〗 𝑑𝜃 = 2 [(sin⁡4 𝜃)/4+𝜃] −4 [(sin⁡2 𝜃)/2]+C Using cos 2x = 1 − 2 sin^2 x 〖𝑠𝑖𝑛〗^2 𝑥=(1 − cos⁡2𝑥)/2 = sin⁡4𝜃/2+2𝜃 −2 𝑠𝑖𝑛 2𝜃+ C Now x = 〖𝑐𝑜𝑠〗^2 2𝜃 √𝑥 " = " cos⁡2𝜃 〖𝑐𝑜𝑠〗^(−1) √𝑥=2𝜃 1/2 〖𝑐𝑜𝑠〗^(−1) √𝑥=𝜃 And, sin 4𝜃 = 2 sin 2𝜃 cos 2𝜃 1 − 𝑥=1−〖𝑐𝑜𝑠〗^2 2𝜃 1 − 𝑥=〖𝑠𝑖𝑛〗^2 2𝜃 sin 2𝜃 = √(1−𝑥) = 2√(1−𝑥)×√𝑥 = 2 √𝑥 √(1−𝑥) Putting the values. = sin⁡4𝜃/2+2θ−2 sin⁡2θ+ C = (2√𝑥 √(1 − 𝑥))/2+2 (〖𝑐𝑜𝑠〗^(−1) √𝑥)/2−2√(1−𝑥)+C = √𝑥 √(1−𝑥)+〖𝑐𝑜𝑠〗^(−1) √𝑥−2√(1−𝑥)+ C = √(𝑥−𝑥^2 )+〖𝑐𝑜𝑠〗^(−1) √𝑥−2√(1−𝑥)+C = –2√(𝟏−𝒙)+〖𝒄𝒐𝒔〗^(−𝟏) √𝒙+√(𝒙−𝒙^𝟐 )+𝐂