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Misc 28 - Chapter 7 Class 12 Integration - Evaluate definite

Misc 28 - Chapter 7 Class 12 Integrals - Part 2
Misc 28 - Chapter 7 Class 12 Integrals - Part 3 Misc 28 - Chapter 7 Class 12 Integrals - Part 4 Misc 28 - Chapter 7 Class 12 Integrals - Part 5 Misc 28 - Chapter 7 Class 12 Integrals - Part 6 Misc 28 - Chapter 7 Class 12 Integrals - Part 7 Misc 28 - Chapter 7 Class 12 Integrals - Part 8 Misc 28 - Chapter 7 Class 12 Integrals - Part 9 Misc 28 - Chapter 7 Class 12 Integrals - Part 10 Misc 28 - Chapter 7 Class 12 Integrals - Part 11

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Misc 27 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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.