Integration Full Chapter Explained - Integration Class 12 - Everything you need



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


Ex 7.11, 10 By using the properties of definite integrals, evaluate the integrals : ∫_0^(πœ‹/2)β–’γ€– (2 log⁑sin⁑π‘₯ βˆ’log⁑sin⁑2π‘₯ ) γ€— 𝑑π‘₯ Let I1=∫_0^(πœ‹/2)β–’γ€– (2 log⁑sin⁑π‘₯ βˆ’log⁑sin⁑2π‘₯ ) γ€— 𝑑π‘₯ I1= ∫_0^(πœ‹/2)β–’γ€– [2 log⁑sin⁑π‘₯ βˆ’π‘™π‘œπ‘”(2 sin⁑〖π‘₯ cos⁑π‘₯ γ€— )] γ€— 𝑑π‘₯ I1= ∫_0^(πœ‹/2)β–’γ€– [2 log⁑sin⁑π‘₯ βˆ’log⁑2βˆ’log⁑sin⁑〖π‘₯βˆ’log⁑cos⁑π‘₯ γ€— ] γ€— 𝑑π‘₯ I1= ∫_0^(πœ‹/2)β–’γ€– [log⁑sin⁑π‘₯ βˆ’π‘™π‘œπ‘”2βˆ’log⁑cos⁑π‘₯ ] γ€— 𝑑π‘₯ I1= ∫_0^(πœ‹/2)β–’log⁑sin⁑〖π‘₯ 𝑑π‘₯γ€— βˆ’βˆ«_0^(πœ‹/2)β–’γ€–log⁑2𝑑π‘₯βˆ’βˆ«_0^(πœ‹/2)β–’log⁑cos⁑〖π‘₯ 𝑑π‘₯γ€— γ€— Solving I2 I2=∫_0^(πœ‹/2)β–’log⁑cos⁑〖π‘₯ 𝑑π‘₯γ€— ∴ I2= ∫_0^(πœ‹/2)β–’logβ‘π‘π‘œπ‘ (πœ‹/2βˆ’π‘₯)𝑑π‘₯ I2=∫_0^(πœ‹/2)β–’log⁑sin⁑〖π‘₯ 𝑑π‘₯γ€— Put the value of I2 in (1) i.e. I1 ∴ I1= ∫_0^(πœ‹/2)β–’log⁑sin⁑〖π‘₯ 𝑑π‘₯γ€— βˆ’βˆ«_0^(πœ‹/2)β–’γ€–log 2〗⁑𝑑π‘₯ βˆ’βˆ«_𝟎^(𝝅/𝟐)β–’π₯π¨π β‘πœπ¨π¬β‘γ€–π’™ 𝒅𝒙〗 I1= ∫_0^(πœ‹/2)β–’log⁑sin⁑〖π‘₯ 𝑑π‘₯γ€— βˆ’βˆ«_0^(πœ‹/2)β–’γ€–log 2〗⁑𝑑π‘₯ βˆ’βˆ«_𝟎^(𝝅/𝟐)β–’π’π’π’ˆβ‘π¬π’π§β‘γ€–π’™ 𝒅𝒙〗 Using the property P4 : ∫_0^π‘Žβ–’γ€–π‘“(π‘₯)𝑑π‘₯=γ€— ∫_0^π‘Žβ–’π‘“(π‘Žβˆ’π‘₯)𝑑π‘₯ I1= – ∫_0^(πœ‹/2)β–’γ€–log 2〗⁑𝑑π‘₯ I1= – log 2∫_0^(πœ‹/2)▒𝑑π‘₯ I1= – log 2[π‘₯]_0^(πœ‹/2) I1= – log 2[πœ‹/2βˆ’0] I1= – log 2Γ—πœ‹/2 I1= log⁑〖〖 (2)γ€—^(βˆ’1) γ€— [πœ‹/2] I1= 𝝅/𝟐 π₯𝐨𝐠 (𝟏/𝟐) (𝐴𝑠 log⁑〖2 𝑖𝑠 π‘π‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘γ€— ) (β–ˆ("Using the property " @π‘Ž log⁑𝑏 = logβ‘π‘π‘Ž))

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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.