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Chapter 7 Class 12 Integrals
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Example 34 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€— 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€— 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 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€— 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 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.