# Misc 28

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Misc 28 Evaluate the definite integral ∫_(π/6)^(π/3)▒〖(sinx + cosx)/√(sin〖2x 〗 ) 〗 ∫_(π/6)^(π/3)▒〖(sinx + cosx)/√(sin〖2x 〗 ) dx 〗 = ∫_(π/6)^(π/3)▒〖 (sinx + cosx)/√(2 sinx cosx ) dx 〗 = ∫_(π/6)^(π/3)▒〖 (sinx/√(2 sinx cosx )+cosx/√(2 sinx cosx )) dx 〗 = ∫_(π/6)^(π/3)▒〖 (√(sinx )/(√2 √(cosx ))+√(cosx )/(√2 √(sinx ))) dx 〗 = ∫_(π/6)^(π/3)▒〖 (1/√2 √(sinx/cosx ) + 1/√2 √(cosx/sinx )) dx 〗 = ∫_(π/6)^(π/3)▒〖 (1/√2 √(sinx/cosx ) + 1/√2 √(cosx/sinx )) dx 〗 = 1/√2 ∫_(π/6)^(π/3)▒〖 (√(tanx )+√(cotx ) ) dx 〗 = 1/√2 ∫1_(π/6)^(π/3)▒〖 [√(cotx )+1/√(cotx )] 〗 dx = 1/√2 ∫1_(π/6)^(π/3)▒〖 [(cotx + 1)/√(cotx )] 〗 dx = 1/√2 ∫1_(π/6)^(π/3)▒〖 [√(tanx ) (cotx+1)] 〗 dx Let tanx=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 ) (cotx+1)] dx = 1/√2 ∫1_(π/6)^(π/3)▒[√(t^2 ) (1/tanx +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)((tanx - 1)/(√2 √(tanx ))) )_(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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Integration Formula Sheet - Chapter 7 Class 12 Formulas Important

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 7 years. He provides courses for Mathematics and Science from Class 6 to 12. You can learn personally from here https://www.teachoo.com/premium/maths-and-science-classes/.