Slide13.JPG

Slide14.JPG
Slide15.JPG Slide16.JPG Slide17.JPG Slide18.JPG Slide19.JPG Slide20.JPG Slide21.JPG Slide22.JPG Slide23.JPG

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Transcript

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)

Ask a doubt
Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

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.