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 9 Find the following integrals: (i) ∫1▒𝑑π‘₯/(π‘₯^2βˆ’ 6π‘₯ + 13) ∫1▒𝑑π‘₯/(π‘₯^2βˆ’ 6π‘₯ + 13) =∫1▒𝑑π‘₯/(π‘₯^2βˆ’ 2 Γ— 3 Γ— π‘₯ + 13) = ∫1▒𝑑π‘₯/((π‘₯^2 βˆ’ 2 . 3 π‘₯ + 3^2 ) + 13 βˆ’ 3^2 ) = ∫1▒𝑑π‘₯/((π‘₯ βˆ’ 3)^2 + 13 βˆ’ 9) = ∫1▒𝑑π‘₯/((π‘₯ βˆ’ 3)^2 + 4) = ∫1▒𝑑π‘₯/((π‘₯ βˆ’ 3)^2 + 2^2 ) ("Adding and subtracting " (3)^2 ) ("Using " π‘Ž^2βˆ’2π‘Žπ‘+𝑏^2=(π‘Žβˆ’π‘)^2 ) It is of form 𝑑π‘₯/(π‘₯^2 + π‘Ž^2 )=1/π‘Ž γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1)⁑〖π‘₯/π‘Žγ€—+𝐢 Replacing π‘₯ with (π‘₯βˆ’3) and π‘Ž with 2 =𝟏/𝟐 〖𝒕𝒂𝒏〗^(βˆ’πŸ)⁑〖(𝒙 βˆ’ πŸ‘)/πŸγ€— +π‘ͺ Example 9 Find the following integrals: (ii) ∫1▒𝑑π‘₯/(γ€–3π‘₯γ€—^2βˆ’13π‘₯ + 10) ∫1▒𝑑π‘₯/(γ€–3π‘₯γ€—^2 βˆ’ 13π‘₯ + 10) Solving denominator γ€–3π‘₯γ€—^2+13π‘₯βˆ’10 =3(π‘₯^2+13/3 π‘₯ βˆ’10/3) =3(π‘₯^2+2. π‘₯Γ— 13/6 βˆ’10/3) Adding and subtracting (13/6)^2 =3(π‘₯^2+2. π‘₯Γ— 13/6+(13/6)^2βˆ’10/3βˆ’(13/6)^2 ) =3((π‘₯+13/6)^2βˆ’10/3βˆ’(169/36)) =3((π‘₯+13/6)^2βˆ’(10/3 +169/36)) =3((π‘₯+13/6)^2βˆ’((120 +169)/36 )) =3((π‘₯+13/6)^2βˆ’289/36) =3((π‘₯+13/6)^2βˆ’(17/6)^2 ) Hence, our equation becomes ∫1▒𝑑π‘₯/(γ€–3π‘₯γ€—^2 βˆ’ 13π‘₯ + 10) = 1/3 ∫1▒𝑑π‘₯/((π‘₯ + 13/6)^2βˆ’ (17/6)^2 ) It is of form ∫1▒〖𝑑π‘₯/(π‘₯^2 βˆ’ π‘Ž^2 )=1/2π‘Ž π‘™π‘œπ‘”|(π‘₯ βˆ’ π‘Ž)/(π‘₯ + π‘Ž)|+𝐢1γ€— Replacing π‘₯ by (π‘₯+13/6)π‘Žπ‘›π‘‘ π‘Ž 𝑏𝑦 17/6, = 1/3 Γ— 1/2(17/6) Γ—log⁑|(π‘₯ + 13/6 βˆ’ 17/6)/(π‘₯+ 13/6 + 17/6)| + C = 1/3 Γ— 6/2(17) Γ—log⁑|((6π‘₯ + 13 βˆ’ 17)/6)/((6π‘₯ +13 + 17)/6)| + C = 1/17 log⁑|(6π‘₯ βˆ’ 4)/(6π‘₯ + 30)| + C = 1/17 log⁑|(2(3π‘₯ βˆ’ 2))/(6(π‘₯ + 5))|+ C = 1/17 log⁑|( (3π‘₯ βˆ’ 2))/(3(π‘₯ + 5))|+ C = 1/17 log⁑|( (3π‘₯ βˆ’ 2))/((π‘₯ + 5))|βˆ’1/17 log⁑3 + C = 𝟏/πŸπŸ• π’π’π’ˆβ‘|( (πŸ‘π’™ βˆ’ 𝟐))/((𝒙 + πŸ“))|+ C1 Example 9 Find the following integrals: (iii) ∫1▒𝑑π‘₯/(√(5π‘₯^2 βˆ’ 2π‘₯) ) ∫1▒𝑑π‘₯/(√(5π‘₯^2 βˆ’ 2π‘₯) ) = ∫1▒𝑑π‘₯/(√(5(π‘₯^2 βˆ’ 2/5 π‘₯) ) ) = ∫1▒𝑑π‘₯/(√(5(π‘₯^2 βˆ’ 2(π‘₯)(1/5)) ) ) = ∫1▒𝑑π‘₯/(√(5(π‘₯^2 βˆ’ 2(π‘₯)(1/5) + (1/5)^2βˆ’ (1/5)^2 ) ) ) = ∫1▒𝑑π‘₯/(√(5[(π‘₯ βˆ’ 1/5)^2βˆ’(1/5)^2 ] ) ) = ∫1▒𝑑π‘₯/(√5 √((π‘₯ βˆ’ 1/5)^2βˆ’(1/5)^2 )) (Taking 5 common) [Adding and subtracting (1/5)^2] = ∫1▒𝑑π‘₯/(√(5[(π‘₯ βˆ’ 1/5)^2βˆ’(1/5)^2 ] ) ) = ∫1▒𝑑π‘₯/(√5 √((π‘₯ βˆ’ 1/5)^2βˆ’(1/5)^2 )) =1/√5 π‘™π‘œπ‘”|π‘₯βˆ’1/5+√((π‘₯βˆ’1/5)^2βˆ’(1/5)^2 )|+𝐢 =1/√5 π‘™π‘œπ‘”|π‘₯βˆ’1/5+√(π‘₯^2+(1/5)^2βˆ’2(π‘₯)(1/5)βˆ’(1/5)^2 )|+𝐢 =𝟏/βˆšπŸ“ π’π’π’ˆ|π’™βˆ’πŸ/πŸ“+√(𝒙^πŸβˆ’πŸπ’™/πŸ“)|+π‘ͺ It is of form ∫1▒〖𝑑π‘₯/(√(π‘₯^2 βˆ’ π‘Ž^2 ) )=π‘™π‘œπ‘”|π‘₯+√(π‘₯^2βˆ’π‘Ž^2 )|+𝐢1γ€— Replacing π‘₯ by (π‘₯βˆ’1/5)π‘Žπ‘›π‘‘ π‘Ž 𝑏𝑦 1/5, (Using√(π‘Ž.𝑏)=βˆšπ‘Ž βˆšπ‘)

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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.