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 _( /6)^( /3) (sin x + cos x)/ (sin 2x ) _( /6)^( /3) (sin x + cos x)/ (sin 2x ) dx = _( /6)^( /3) (sin x + cos x)/ (2 sin x cos x ) dx = _( /6)^( /3) (sin x/ (2 sin x cos x )+cos x/ (2 sin x cos x )) dx = _( /6)^( /3) ( (sin x )/( 2 (cos x ))+ (cos x )/( 2 (sin x ))) dx = _( /6)^( /3) (1/ 2 (sin x/cos x ) + 1/ 2 (cos x/sin x )) dx = _( /6)^( /3) (1/ 2 (sin x/cos x ) + 1/ 2 (cos x/sin x )) dx = 1/ 2 _( /6)^( /3) ( (tan x )+ (cot x ) ) dx = 1/ 2 1_( /6)^( /3) [ (cot x )+1/ (cot x )] dx = 1/ 2 1_( /6)^( /3) [(cot x + 1)/ (cot x )] dx = 1/ 2 1_( /6)^( /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_( /6)^( /3) [ (t^2 ) (cot x+1)] dx = 1/ 2 1_( /6)^( /3) [ (t^2 ) (1/tan x +1)] dx = 1/ 2 1_( /6)^( /3) t[1/t^2 +1] dx = 1/ 2 1_( /6)^( /3) t[1/t^2 +1] dx = 1/ 2 1_( /6)^( /3) t[(1 + t^2)/t^2 ] dx = 1/ 2 1_( /6)^( /3) t[(1 + t^2)/t^2 ] 2t/(1 + t^2 ) . dt = 1/ 2 1_( /6)^( /3) 2[(1 + t^2)/(1 + t^4 )] dt = 1/ 2 2 1_( /6)^( /3) (1 + t^2)/(1 + t^4 ) dt Dividing numerator and denominator by t^2 = 2 1_( /6)^( /3) ((1 + t^2)/t^2 )/((1 + t^4)/t^2 ) . dt = 2 1_( /6)^( /3) (1/t^2 + 1)/(1/t^2 + t^2 ) . dt = 2 1_( /6)^( /3) (1 + 1/t^2 )/( t^2 + 1/t^2 + 2 - 2) . dt = 2 1_( /6)^( /3) (1 + 1/t^2 )/( (t)^2 + (1/t)^2- 2 (t) (1/t) + 2) . dt = 2 1_( /6)^( /3) (1 + 1/t^2 )/((t - 1/t)^2 + 2) . dt = 2 1_( /6)^( /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_( /6)^( /3) (1 + 1/t^2 )/(y^2 +( 2 )^2 ) . dt = 2 1_( /6)^( /3) ((1 + 1/t^2 ))/(y^2 +( 2 )^2 ) dy/((1 - 1/t^2 ) ) = 2 1_( /6)^( /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 )_( /6)^( /3) = (tan^(-1) y/ 2 )_( /6)^( /3) = (tan^(-1) (1/t - t)/ 2 )_( /6)^( /3) = (tan^(-1) (t^2 - 1)/( 2 t) )_( /6)^( /3) = (tan^(-1) ((tan x - 1)/( 2 (tan x ))) )_( /6)^( /3) = tan^(-1) ((tan ( /3) - 1)/ (2 tan ( /3) ))-tan^(-1) ((tan ( /6) - 1)/ (2 tan ( /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 = BC/AB sin = ( 3 - 1)/2 = sin ^(-1) (( 3 - 1)/2)

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.