Check sibling questions

Example 36 - Integration of log (sin x) from 0 to pi/2 - Teachoo

Example 36 - Chapter 7 Class 12 Integrals - Part 2
Example 36 - Chapter 7 Class 12 Integrals - Part 3 Example 36 - Chapter 7 Class 12 Integrals - Part 4 Example 36 - Chapter 7 Class 12 Integrals - Part 5

Get live Maths 1-on-1 Classs - Class 6 to 12


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

Ask a doubt
Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

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.