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Ex 7.11, 10 - Evaluate (2 log sin x - log sin 2x) dx - Ex 7.11

Ex 7.11, 10 - Chapter 7 Class 12 Integrals - Part 2
Ex 7.11, 10 - Chapter 7 Class 12 Integrals - Part 3

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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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.