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 36 Evaluate ∫_0^(πœ‹/2 )β–’log⁑sin⁑π‘₯ 𝑑π‘₯ Let I1=∫_0^(πœ‹/2 )β–’π‘™π‘œπ‘”(𝑠𝑖𝑛π‘₯) 𝑑π‘₯ ∴ I1=∫_0^(πœ‹/2)▒𝑠𝑖𝑛(πœ‹/2βˆ’π‘₯)𝑑π‘₯ I1= ∫_0^(πœ‹/2)β–’π‘™π‘œπ‘”(cos⁑π‘₯ )𝑑π‘₯ Adding (1) and (2) i.e. (1) + (2) I1+ I1=∫_0^(πœ‹/2)β–’γ€–π‘™π‘œπ‘”(sin⁑π‘₯ )𝑑π‘₯+∫_0^(πœ‹/2)β–’π‘™π‘œπ‘”(cos⁑π‘₯ )𝑑π‘₯γ€— Using Property P4 ∫_0^π‘Žβ–’γ€–π‘“(π‘₯)𝑑π‘₯=γ€— ∫_0^π‘Žβ–’π‘“(π‘Žβˆ’π‘₯)𝑑π‘₯ …(1) …(2) 2I1 =∫_0^(πœ‹/2)β–’γ€–log⁑[sin⁑〖π‘₯ cos⁑π‘₯ γ€— ] 𝑑π‘₯γ€— 2I1 = ∫_0^(πœ‹/2)β–’γ€–log⁑[2sin⁑〖π‘₯ cos⁑π‘₯ γ€—/2] 𝑑π‘₯γ€— 2I1 = ∫_0^(πœ‹/2)β–’[log[2sin⁑〖π‘₯ cos⁑π‘₯ γ€— ]βˆ’log⁑2 ]𝑑π‘₯ 2I1 = ∫_0^(πœ‹/2)β–’[log[sin⁑2π‘₯ ]βˆ’log⁑2 ]𝑑π‘₯ 2I1 = ∫_0^(πœ‹/2)β–’log[sin⁑2π‘₯ ]𝑑π‘₯βˆ’βˆ«_0^(πœ‹/2)β–’γ€–log 2 𝑑π‘₯γ€— Solving 𝐈𝟐 I2=∫_0^(πœ‹/2)β–’γ€–log sin⁑2π‘₯ 𝑑π‘₯γ€— "(Using" logβ‘π‘Ž + log⁑𝑏 = log⁑(π‘Ž.𝑏)) ("Using " log(π‘Ž/𝑏) = log⁑(π‘Ž) – log⁑(𝑏)) ("Using" sin⁑2π‘₯=2 sin⁑〖π‘₯ cos⁑π‘₯ γ€—) Let 2π‘₯=𝑑 Differentiating both sides w.r.t.π‘₯ 2=𝑑𝑑/𝑑π‘₯ 𝑑π‘₯=𝑑𝑑/2 ∴ Putting the values of t and 𝑑𝑑 and changing the limits, I2 =∫_0^(πœ‹/2)β–’log(sin⁑2π‘₯ )𝑑π‘₯ I2 = ∫_0^πœ‹β–’γ€–log(sin⁑𝑑 ) 𝑑𝑑/2γ€— I2 = 1/2 ∫_0^πœ‹β–’log(sin⁑𝑑 )𝑑𝑑 Using the Property, P6 :- ∫_0^2π‘Žβ–’π‘“(π‘₯)𝑑π‘₯=2∫_0^π‘Žβ–’π‘“(π‘₯)𝑑π‘₯, 𝑖𝑓 𝑓(2π‘Žβˆ’π‘₯)=𝑓(π‘₯) Here, 𝑓(𝑑)=log⁑𝑠𝑖𝑛𝑑 𝑓(2π‘Žβˆ’π‘‘)=𝑓(2πœ‹βˆ’π‘‘)=log⁑𝑠𝑖𝑛(2πœ‹βˆ’π‘‘)=log⁑sin⁑𝑑 Since 𝑓(𝑑)=𝑓(2π‘Žβˆ’π‘‘) ∴ I2 = 1/2 ∫_0^πœ‹β–’log⁑sin⁑〖𝑑 𝑑𝑑〗 =1/2 Γ—2∫_0^(πœ‹/2)β–’log⁑sin⁑〖𝑑. 𝑑𝑑〗 =∫_0^(πœ‹/2)β–’log⁑sin⁑〖𝑑. 𝑑𝑑〗 I2=∫_0^(πœ‹/2)β–’log⁑sin⁑〖π‘₯ 𝑑π‘₯γ€— Now, Using the Property, P0 :- ∫_π‘Ž^𝑏▒𝑓(π‘₯)𝑑π‘₯=∫_π‘Ž^𝑏▒𝑓(𝑑)𝑑𝑑 Putting the value of I2 in equation (3), we get 2I1 =∫_𝟎^(𝝅/𝟐)β–’π₯𝐨𝐠[π’”π’Šπ’β‘πŸπ’™ ]⁑𝒅𝒙 βˆ’βˆ«_0^(πœ‹/2)β–’log(2)⁑𝑑π‘₯ 2I1 = ∫_𝟎^(𝝅/𝟐)β–’π₯𝐨𝐠(π’”π’Šπ’β‘π’™ )⁑𝒅𝒙 βˆ’log(2) ∫_0^(πœ‹/2)β–’γ€–1.〗⁑𝑑π‘₯ 2I1 = 𝐈𝟏 βˆ’ log(2) [π‘₯]_0^(πœ‹/2) 2I1βˆ’I1=βˆ’log⁑2 [πœ‹/2βˆ’0] I1=βˆ’log⁑2 [πœ‹/2] ∴ 𝐈𝟏=(βˆ’ 𝝅)/𝟐 π₯𝐨𝐠⁑𝟐

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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.