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Chapter 7 Class 12 Integrals
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Ex 7.10, 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〗 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)

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.