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Misc 28 - Chapter 7 Class 12 Integration - Evaluate definite - Miscellaneous

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  1. Chapter 7 Class 12 Integrals
  2. Serial order wise
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Misc 28 Evaluate the definite integral ∫_(π/6)^(π/3)▒〖(sin⁡x + cos⁡x)/√(sin⁡〖2x 〗 ) 〗 ∫_(π/6)^(π/3)▒〖(sin⁡x + cos⁡x)/√(sin⁡〖2x 〗 ) dx 〗 = ∫_(π/6)^(π/3)▒〖 (sin⁡x + cos⁡x)/√(2 sin⁡x cos⁡x ) dx 〗 = ∫_(π/6)^(π/3)▒〖 (sin⁡x/√(2 sin⁡x cos⁡x )+cos⁡x/√(2 sin⁡x cos⁡x )) dx 〗 = ∫_(π/6)^(π/3)▒〖 (√(sin⁡x )/(√2 √(cos⁡x ))+√(cos⁡x )/(√2 √(sin⁡x ))) dx 〗 = ∫_(π/6)^(π/3)▒〖 (1/√2 √(sin⁡x/cos⁡x ) + 1/√2 √(cos⁡x/sin⁡x )) dx 〗 = ∫_(π/6)^(π/3)▒〖 (1/√2 √(sin⁡x/cos⁡x ) + 1/√2 √(cos⁡x/sin⁡x )) dx 〗 = 1/√2 ∫_(π/6)^(π/3)▒〖 (√(tan⁡x )+√(cot⁡x ) ) dx 〗 = 1/√2 ∫1_(π/6)^(π/3)▒〖 [√(cot⁡x )+1/√(cot⁡x )] 〗 dx = 1/√2 ∫1_(π/6)^(π/3)▒〖 [(cot⁡x + 1)/√(cot⁡x )] 〗 dx = 1/√2 ∫1_(π/6)^(π/3)▒〖 [√(tan⁡x ) (cot⁡x+1)] 〗 dx Let tan⁡x=t^2 Differentiating both sides w.r.t.x. sec^2 x=2t dt/dx 1+tan^2 x=2t . dt/dx 1+(t^2 )^2=2t . dt/dx 1+t^4=2t . dt/dx (1+t^4 ) dx=2t dt dx=2t/(1 + t^4 ) . dt Putting values of t & dt, we get 1/√2 ∫1_(π/6)^(π/3)▒[√(t^2 ) (cot⁡x+1)] dx = 1/√2 ∫1_(π/6)^(π/3)▒[√(t^2 ) (1/tan⁡x +1)] dx = 1/√2 ∫1_(π/6)^(π/3)▒t[1/t^2 +1] dx = 1/√2 ∫1_(π/6)^(π/3)▒t[1/t^2 +1] dx = 1/√2 ∫1_(π/6)^(π/3)▒t[(1 + t^2)/t^2 ] dx = 1/√2 ∫1_(π/6)^(π/3)▒t[(1 + t^2)/t^2 ] ×2t/(1 + t^2 ) . dt = 1/√2 ∫1_(π/6)^(π/3)▒2[(1 + t^2)/(1 + t^4 )] dt = 1/√2 2∫1_(π/6)^(π/3)▒(1 + t^2)/(1 + t^4 ) dt Dividing numerator and denominator by t^2 = √2 ∫1_(π/6)^(π/3)▒〖 ((1 + t^2)/t^2 )/((1 + t^4)/t^2 )〗. dt = √2 ∫1_(π/6)^(π/3)▒〖 (1/t^2 + 1)/(1/t^2 + t^2 )〗. dt = √2 ∫1_(π/6)^(π/3)▒〖 (1 + 1/t^2 )/( t^2 + 1/t^2 + 2 - 2)〗. dt = √2 ∫1_(π/6)^(π/3)▒〖 (1 + 1/t^2 )/( (t)^2 + (1/t)^2- 2 (t) (1/t) + 2)〗. dt = √2 ∫1_(π/6)^(π/3)▒〖 (1 + 1/t^2 )/((t - 1/t)^2 + 2)〗. dt = √2 ∫1_(π/6)^(π/3)▒〖 (1 + 1/t^2 )/((t - 1/t)^2 +(√2 )^2 )〗. dt Let t-1/t=y Differentiating both sides w.r.t.x. 1+ 1/t^2 = dy/dt dt =dy/((1 + 1/t^2 ) ) Putting the values of (1/t -t) and dt, we get = √2 ∫1_(π/6)^(π/3)▒〖 (1 + 1/t^2 )/(y^2 +(√2 )^2 )〗. dt = √2 ∫1_(π/6)^(π/3)▒〖 ((1 + 1/t^2 ))/(y^2 +(√2 )^2 )〗× dy/((1 - 1/t^2 ) ) = √2 ∫1_(π/6)^(π/3)▒〖 1/(y^2 +(√2 )^2 )〗. dy It is of form ∫1▒1/(x^2 +〖 a〗^2 ) . dx = 1/a tan^(-1)⁡〖 x/a〗 +C1 Replacing x by y and a by √2 we, get = √2 (1/√2 tan^(-1)⁡〖 y/√2〗 )_(π/6)^(π/3) = (tan^(-1)⁡〖 y/√2〗 )_(π/6)^(π/3) = (tan^(-1)⁡〖 (1/t - t)/√2〗 )_(π/6)^(π/3) = (tan^(-1)⁡〖 (t^2 - 1)/(√2 t)〗 )_(π/6)^(π/3) = (tan^(-1)⁡((tan⁡x - 1)/(√2 √(tan⁡x ))) )_(π/6)^(π/3) = tan^(-1) ((tan⁡(π/3) - 1)/√(2 tan⁡(π/3) ))-tan^(-1) ((tan⁡(π/6) - 1)/√(2 tan⁡(π/6) )) = tan^(-1) ((√3 - 1)/√(2 √3) )-tan^(-1) ((1 - √3 )/(√3 √(2 . 1/(√3 " " )))) = tan^(-1) ((√3 - 1)/√(2 √3) )-tan^(-1) ((1 - √3 )/√(3 . 2 . 1/(√3 " " ))) = tan^(-1) ((√3 - 1)/√(2 √3) )-tan^(-1) ((1 - √3 )/√(2 √3) ) = tan^(-1) ((√3 - 1)/√(2 √3) )-tan^(-1) ((-(√3 - 1))/√(2 √3) ) = tan^(-1) ((√3 - 1)/√(2 √3) )+tan^(-1) ((√3 - 1)/√(2 √3) ) = 2 tan^(-1) ((√3 - 1)/√(2 √3) ) = 2 〖sin〗^(-1) [(√3 - 1)/2] Rough AB^2=BC^2+AC^2 AB^2=(√3-1)^2+(√(2 √3) )^2 AB^2=3+1-2 √3+2 √3 AB^2=4 ∴ AB=2 sin 𝜃 = BC/AB sin 𝜃 = (√3 - 1)/2 𝜃 = 〖sin〗^(-1) ((√3 - 1)/2)

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
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