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  1. Chapter 7 Class 12 Integrals
  2. Serial order wise

Transcript

Misc 28 Evaluate the definite integral ∫_(πœ‹/6)^(πœ‹/3)β–’γ€–(sin⁑π‘₯ + cos⁑π‘₯)/√(sin⁑〖2π‘₯ γ€— ) γ€— ∫_(πœ‹/6)^(πœ‹/3)β–’γ€–(sin⁑π‘₯ + cos⁑π‘₯)/√(sin⁑〖2π‘₯ γ€— ) 𝑑π‘₯ γ€— = ∫_(πœ‹/6)^(πœ‹/3)β–’γ€– (sin⁑π‘₯ + cos⁑π‘₯)/√(2 sin⁑π‘₯ cos⁑π‘₯ ) 𝑑π‘₯ γ€— = ∫_(πœ‹/6)^(πœ‹/3)β–’γ€– (sin⁑π‘₯/√(2 sin⁑π‘₯ cos⁑π‘₯ )+cos⁑π‘₯/√(2 sin⁑π‘₯ cos⁑π‘₯ )) 𝑑π‘₯ γ€— = ∫_(πœ‹/6)^(πœ‹/3)β–’γ€– (√(sin⁑π‘₯ )/(√2 √(cos⁑π‘₯ ))+√(cos⁑π‘₯ )/(√2 √(sin⁑π‘₯ ))) 𝑑π‘₯ γ€— ("Using" sin⁑2πœƒ=2 sinβ‘πœƒ cosβ‘πœƒ ) = ∫_(πœ‹/6)^(πœ‹/3)β–’γ€– (1/√2 √(sin⁑π‘₯/cos⁑π‘₯ ) + 1/√2 √(cos⁑π‘₯/sin⁑π‘₯ )) 𝑑π‘₯ γ€— = 1/√2 ∫_(πœ‹/6)^(πœ‹/3)β–’γ€–(√(tan⁑π‘₯ )+√(cot⁑π‘₯ ) ) 𝑑π‘₯ γ€— = 1/√2 ∫1_(πœ‹/6)^(πœ‹/3)β–’γ€– [√(cot⁑π‘₯ )+1/√(cot⁑π‘₯ )] γ€— 𝑑π‘₯ = 1/√2 ∫1_(πœ‹/6)^(πœ‹/3)β–’(cot⁑π‘₯ + 1)/√(cot⁑π‘₯ ) 𝑑π‘₯ = 1/√2 ∫1_(πœ‹/6)^(πœ‹/3)β–’γ€–βˆš(tan⁑π‘₯ ) (cot⁑π‘₯+1) γ€— 𝑑π‘₯ ("Using" π‘‘π‘Žπ‘›β‘π‘₯= 1/π‘π‘œπ‘‘β‘π‘₯ ) Let tan⁑π‘₯=𝑑^2 Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯. sec^2 π‘₯=2𝑑 𝑑𝑑/𝑑π‘₯ 1+tan^2 π‘₯=2𝑑 . 𝑑𝑑/𝑑π‘₯ 1+(𝑑^2 )^2=2𝑑 . 𝑑𝑑/𝑑π‘₯ 1+𝑑^4=2𝑑 . 𝑑𝑑/𝑑π‘₯ (1+𝑑^4 ) 𝑑π‘₯=2𝑑 𝑑𝑑 𝑑π‘₯=2𝑑/(1 + 𝑑^4 ) 𝑑𝑑 Putting values of t & dt, we get ("Using" π‘‘π‘Žπ‘›β‘π‘₯= 𝑑^2) Putting values of t & dt, we get 1/√2 ∫1_(πœ‹/6)^(πœ‹/3)β–’γ€–βˆš(tan⁑π‘₯ ) (cot⁑π‘₯+1) γ€— 𝑑π‘₯ =1/√2 ∫1_(πœ‹/6)^(πœ‹/3)β–’[√(𝑑^2 ) (cot⁑π‘₯+1)] 𝑑π‘₯ = 1/√2 ∫1_(πœ‹/6)^(πœ‹/3)β–’[√(𝑑^2 ) (1/tan⁑π‘₯ +1)] 𝑑π‘₯ = 1/√2 ∫1_(πœ‹/6)^(πœ‹/3)▒𝑑[1/𝑑^2 +1] 𝑑π‘₯ = 1/√2 ∫1_(πœ‹/6)^(πœ‹/3)▒𝑑[(1 + 𝑑^2)/𝑑^2 ] 𝑑π‘₯ = 1/√2 ∫1_(πœ‹/6)^(πœ‹/3)▒𝑑[(1 + 𝑑^2)/𝑑^2 ] Γ—2𝑑/(1 + 𝑑^2 ) . 𝑑𝑑 = 1/√2 ∫1_(πœ‹/6)^(πœ‹/3)β–’2[(1 + 𝑑^2)/(1 + 𝑑^4 )] 𝑑𝑑 = 1/√2 2∫1_(πœ‹/6)^(πœ‹/3)β–’(1 + 𝑑^2)/(1 + 𝑑^4 ) 𝑑𝑑 Dividing numerator and denominator by 𝑑^2 = √2 ∫1_(πœ‹/6)^(πœ‹/3)β–’γ€– ((1 + 𝑑^2)/𝑑^2 )/((1 + 𝑑^4)/𝑑^2 )γ€—. 𝑑𝑑 = √2 ∫1_(πœ‹/6)^(πœ‹/3)β–’γ€– (1/𝑑^2 + 1)/(1/𝑑^2 + 𝑑^2 )γ€—. 𝑑𝑑 = √2 ∫1_(πœ‹/6)^(πœ‹/3)β–’(1 + 1/𝑑^2 )/( 𝑑^2 + 1/𝑑^2 + 2 βˆ’ 2). 𝑑𝑑 = √2 ∫1_(πœ‹/6)^(πœ‹/3)β–’(1 + 1/𝑑^2 )/( (𝑑)^2 + (1/𝑑)^2βˆ’ 2 (𝑑) (1/𝑑) + 2). 𝑑𝑑 = √2 ∫1_(πœ‹/6)^(πœ‹/3)β–’γ€– (1 + 1/𝑑^2 )/((𝑑 βˆ’ 1/𝑑)^2 + 2)γ€—. 𝑑𝑑 = √2 ∫1_(πœ‹/6)^(πœ‹/3)β–’(1 + 1/𝑑^2 )/((𝑑 βˆ’ 1/𝑑)^2 +(√2 )^2 ). 𝑑𝑑 (Adding and subtracting 2 in denominator) Let π‘‘βˆ’1/𝑑=𝑦 Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯. 1+ 1/𝑑^2 = 𝑑𝑦/𝑑𝑑 𝑑𝑑 =𝑑𝑦/((1 + 1/𝑑^2 ) ) Putting the values of (1/t βˆ’t) and dt, we get √2 ∫1_(πœ‹/6)^(πœ‹/3)β–’(1 + 1/𝑑^2 )/((𝑑 βˆ’ 1/𝑑)^2 +(√2 )^2 ). 𝑑𝑑 = √2 ∫1_(πœ‹/6)^(πœ‹/3)β–’γ€– (1 + 1/𝑑^2 )/(𝑦^2 +(√2 )^2 )γ€—. 𝑑𝑑 = √2 ∫1_(πœ‹/6)^(πœ‹/3)β–’γ€– ((1 + 1/𝑑^2 ))/(𝑦^2 +(√2 )^2 )γ€—Γ— 𝑑𝑦/((1 βˆ’ 1/𝑑^2 ) ) = √2 ∫1_(πœ‹/6)^(πœ‹/3)β–’γ€– 1/(𝑦^2 +(√2 )^2 )γ€—. 𝑑𝑦 = √2 (1/√2 tan^(βˆ’1)⁑〖 𝑦/√2γ€— )_(πœ‹/6)^(πœ‹/3) = (tan^(βˆ’1)⁑〖 𝑦/√2γ€— )_(πœ‹/6)^(πœ‹/3) = (tan^(βˆ’1)⁑〖 (1/𝑑 βˆ’ 𝑑)/√2γ€— )_(πœ‹/6)^(πœ‹/3) ["Using" 𝑦=1/𝑑 βˆ’π‘‘] = (tan^(βˆ’1)⁑〖 (𝑑^2 βˆ’ 1)/(√2 𝑑)γ€— )_(πœ‹/6)^(πœ‹/3) = (tan^(βˆ’1)⁑((tan⁑π‘₯ βˆ’ 1)/(√2 √(tan⁑π‘₯ ))) )_(πœ‹/6)^(πœ‹/3) = tan^(βˆ’1) ((tan⁑(πœ‹/3) βˆ’ 1)/√(2 tan⁑(πœ‹/3) ))βˆ’tan^(βˆ’1) ((tan⁑(πœ‹/6) βˆ’ 1)/√(2 tan⁑(πœ‹/6) )) = tan^(βˆ’1) ((√3 βˆ’ 1)/√(2 √3) )βˆ’tan^(βˆ’1) ((1/√3 βˆ’1 )/(√3 √(2 . 1/(√3 " " )))) = tan^(βˆ’1) ((√3 βˆ’ 1)/√(2 √3) )βˆ’tan^(βˆ’1) ((1 βˆ’ √3 )/(√3 √(2 . 1/(√3 " " )))) Using π‘‘π‘Žπ‘›β‘(πœ‹/3) = √3 π‘‘π‘Žπ‘›β‘(πœ‹/6) = 1/(√3 " " ) = tan^(βˆ’1) ((√3 βˆ’ 1)/√(2 √3) )βˆ’tan^(βˆ’1) ((1 βˆ’ √3 )/√(3 . 2 . 1/(√3 " " ))) = tan^(βˆ’1) ((√3 βˆ’ 1)/√(2 √3) )βˆ’tan^(βˆ’1) ((1 βˆ’ √3 )/√(2 √3) ) = tan^(βˆ’1) ((√3 βˆ’ 1)/√(2 √3) )βˆ’tan^(βˆ’1) ((βˆ’(√3 βˆ’ 1))/√(2 √3) ) = tan^(βˆ’1) ((√3 βˆ’ 1)/√(2 √3) )+tan^(βˆ’1) ((√3 βˆ’ 1)/√(2 √3) ) = 2 tan^(βˆ’1) ((√3 βˆ’ 1)/√(2 √3) ) = 𝟐 〖𝐬𝐒𝒏〗^(βˆ’πŸ) [(βˆšπŸ‘ βˆ’ 𝟏)/𝟐] "Using " tan^(βˆ’1)⁑(βˆ’πœƒ) =βˆ’tan^(βˆ’1)⁑(πœƒ) Note 𝐴𝐡^2=𝐡𝐢^2+𝐴𝐢^2 𝐴𝐡^2=(√3βˆ’1)^2+(√(2 √3) )^2 𝐴𝐡^2=3+1βˆ’2 √3+2 √3 𝐴𝐡^2=4 ∴ 𝐴𝐡=2 sin πœƒ = 𝐡𝐢/𝐴𝐡 sin πœƒ = (√3 βˆ’ 1)/2 πœƒ = 〖𝑠𝑖𝑛〗^(βˆ’1) ((√3 βˆ’ 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 9 years. He provides courses for Maths and Science at Teachoo.