Integration Full Chapter Explained - https://you.tube/Integration-Class-12

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

Transcript

Example 41 Evaluate ∫1β–’[√(cot⁑π‘₯ )+√(tan⁑π‘₯ )] 𝑑π‘₯ ∫1β–’[√(cot⁑π‘₯ )+√(tan⁑π‘₯ )] 𝑑π‘₯ =∫1β–’[√(cot⁑π‘₯ )+1/√(cot⁑π‘₯ )] 𝑑π‘₯ =∫1β–’[(cot⁑π‘₯ + 1)/√(cot⁑π‘₯ )] 𝑑π‘₯ =∫1β–’[√(tan⁑π‘₯ ) (cot⁑π‘₯+1)] 𝑑π‘₯ Let tan⁑π‘₯=𝑑^2 ∫1β–’[√(cot⁑π‘₯ )+√(tan⁑π‘₯ )] 𝑑π‘₯ =∫1β–’[√(cot⁑π‘₯ )+1/√(cot⁑π‘₯ )] 𝑑π‘₯ =∫1β–’[(cot⁑π‘₯ + 1)/√(cot⁑π‘₯ )] 𝑑π‘₯ =∫1β–’[√(tan⁑π‘₯ ) (cot⁑π‘₯+1)] 𝑑π‘₯ Let tan⁑π‘₯=𝑑^2 ("Using" π‘‘π‘Žπ‘›β‘π‘₯= 1/π‘π‘œπ‘‘β‘π‘₯ ) 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 ∫1β–’[√(tan⁑π‘₯ ) (cot⁑π‘₯+1)] 𝑑π‘₯ = ∫1β–’[√(𝑑^2 ) (cot⁑π‘₯+1)] 𝑑π‘₯ ["Using" π‘‘π‘Žπ‘›β‘π‘₯= 𝑑^2 ] = ∫1β–’[√(𝑑^2 ) (1/tan⁑π‘₯ +1)] 𝑑π‘₯ = ∫1▒𝑑[1/𝑑^2 +1] 𝑑π‘₯ = ∫1▒𝑑[(1 + 𝑑^2)/𝑑^2 ] Γ—2𝑑/(1 + 𝑑^4 ) . 𝑑𝑑 = ∫1β–’2[(1 + 𝑑^2)/(1 + 𝑑^4 )] 𝑑𝑑 = 2∫1β–’(1 + 𝑑^2)/(1 + 𝑑^4 ) 𝑑𝑑 Dividing numerator and denominator by 𝑑^2 = 2 ∫1β–’((1 + 𝑑^2)/𝑑^2 )/((1 + 𝑑^4)/𝑑^2 ) . 𝑑𝑑 ["Using" π‘‘π‘Žπ‘›β‘π‘₯= 𝑑^2 ] = 2 ∫1β–’(1/𝑑^2 + 1)/(1/𝑑^2 + 𝑑^2 ) . 𝑑𝑑 = 2 ∫1β–’(1/𝑑^2 + 1)/(1/𝑑^2 + 𝑑^2 ) . 𝑑𝑑 = 2 ∫1β–’(1 + 1/𝑑^2 )/( 𝑑^2 + 1/𝑑^2 + 2 βˆ’ 2) . 𝑑𝑑 = 2 ∫1β–’(1 + 1/𝑑^2 )/( (𝑑)^2 + (1/𝑑)^2βˆ’ 2 (𝑑) (1/𝑑) + 2) . 𝑑𝑑 = 2 ∫1β–’(1 + 1/𝑑^2 )/((𝑑 βˆ’ 1/𝑑)^2 + 2) . 𝑑𝑑 (Adding and subtracting 2 in denominator) = 2 ∫1β–’(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 ∫1β–’(1 + 1/𝑑^2 )/(𝑦^2 +(√2 )^2 ) . 𝑑𝑑 = 2 ∫1β–’((1 + 1/𝑑^2 ))/(𝑦^2 + (√2 )^2 ) Γ— 𝑑𝑦/((1 + 1/𝑑^2 ) ) = 2 ∫1β–’1/(𝑦^2 + (√2 )^2 ) . 𝑑𝑦 = 2(1/√2 tan^(βˆ’1)⁑〖 𝑦/√2γ€— +𝐢1) It is of form ∫1β–’1/(π‘₯^2 +γ€– π‘Žγ€—^2 ) . 𝑑π‘₯ = 1/π‘Ž tan^(βˆ’1)⁑〖 π‘₯/π‘Žγ€— +𝐢1 Replacing x by y and a by √2 we, get = 2/√2 tan^(βˆ’1)⁑〖 𝑦/√2γ€— +2𝐢1 = √2 tan^(βˆ’1)⁑〖 (1/𝑑 βˆ’ 𝑑)/√2γ€— +2𝐢1 = √2 tan^(βˆ’1)⁑〖 (𝑑^2 βˆ’ 1)/(√2 𝑑)γ€— +𝐢 = √2 tan^(βˆ’1)⁑((tan⁑π‘₯ βˆ’ 1)/(√2 √(tan⁑π‘₯ )))+𝐢 = √𝟐 〖𝒕𝒂𝒏〗^(βˆ’πŸ)⁑((𝒕𝒂𝒏⁑𝒙 βˆ’ 𝟏)/(√(𝟐 𝒕𝒂𝒏⁑𝒙 ) ))+π‘ͺ ["Using" 𝑦=1/𝑑 βˆ’π‘‘] [β–ˆ("Using" 𝑑= √(tan⁑π‘₯ ))]

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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.