Chapter 7 Class 12 Integrals
Concept wise

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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€— = ∫_(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‘₯ 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 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 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) = (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 " " )))) = 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] 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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.