# 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)

Misc 1

Misc 2

Misc 3

Misc 4

Misc 5

Misc 6

Misc 7

Misc 8 Important

Misc 9

Misc 10

Misc 11

Misc 12

Misc 13

Misc 14

Misc 15

Misc 16

Misc 17

Misc 18 Important

Misc 19 Important

Misc 20 Important

Misc 21

Misc 22

Misc 23

Misc 24 Important

Misc 25

Misc 26

Misc 27

Misc 28 You are here

Misc 29

Misc 30 Important

Misc 31

Misc 32 Important

Misc 33

Misc 34

Misc 35

Misc 36

Misc 37

Misc 38

Misc 39

Misc 40

Misc 41 Important

Misc 42

Misc 43

Misc 44 Important

Integration Formula Sheet - Chapter 7 Class 12 Formulas Important

About the Author

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.