Slide40.JPG Slide41.JPG Slide42.JPG Slide43.JPG Slide44.JPG Slide45.JPG Slide46.JPG

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Ex 7.10, 16 By using the properties of definite integrals, evaluate the integrals : ∫_0^πœ‹β–’log⁑(1+cos⁑π‘₯ ) 𝑑π‘₯ Let I=∫_𝟎^π…β–’π’π’π’ˆβ‘(𝟏+𝒄𝒐𝒔⁑𝒙 ) 𝒅𝒙 ∴ I=∫_0^πœ‹β–’log⁑(1+π‘π‘œπ‘ (πœ‹βˆ’π‘₯)) 𝑑π‘₯ 𝐈=∫_𝟎^𝝅▒π₯𝐨𝐠⁑(πŸβˆ’π’„π’π’”β‘π’™ ) 𝑑π‘₯ Adding (1) and (2) i.e. (1) + (2) I+I=∫_0^πœ‹β–’π‘™π‘œπ‘”(1+cos⁑π‘₯ )𝑑π‘₯+∫_0^πœ‹β–’π‘™π‘œπ‘”(1βˆ’cos⁑π‘₯ )𝑑π‘₯ 𝟐𝐈=∫_𝟎^𝝅▒[π’π’π’ˆ(𝟏+𝒄𝒐𝒔⁑𝒙 )+π’π’π’ˆ(πŸβˆ’π’„π’π’”β‘π’™ )]𝒅𝒙 "Using log(a) + log(b)" = "log(a.b)") 2I=∫_0^πœ‹β–’π’π’π’ˆ[(𝟏+𝒄𝒐𝒔⁑𝒙 )(πŸβˆ’π’„π’π’”β‘π’™ )]𝒅𝒙 2I=∫_0^πœ‹β–’π‘™π‘œπ‘”[1βˆ’cos^2⁑π‘₯ ]𝑑π‘₯ 2I=∫_0^πœ‹β–’π‘™π‘œπ‘”(〖𝑠𝑖𝑛〗^2⁑π‘₯ )𝑑π‘₯ Using log 𝒂^𝒃=𝒃 π’π’π’ˆβ‘π’‚ 2I=∫_0^πœ‹β–’γ€–2 π‘™π‘œπ‘”(𝑠𝑖𝑛⁑π‘₯ )𝑑π‘₯γ€— 2I=2∫_0^πœ‹β–’π‘™π‘œπ‘”(𝑠𝑖𝑛⁑π‘₯ )𝑑π‘₯ 𝐈=∫_𝟎^π…β–’π’π’π’ˆ(π’”π’Šπ’β‘π’™ )𝒅𝒙 Here, 𝒇(𝒙)=log⁑sin⁑π‘₯ f(π…βˆ’π’™)=π‘™π‘œπ‘”[𝑠𝑖𝑛(πœ‹βˆ’π‘₯)]𝑑π‘₯ =π‘™π‘œπ‘”(sin⁑π‘₯ )𝑑π‘₯ =𝒇(𝒙) Therefore, I=∫_𝟎^π…β–’π’π’π’ˆ(𝐬𝐒𝐧⁑𝒙 )𝒅𝒙=𝟐∫_𝟎^(𝝅/𝟐)β–’π’π’π’ˆ(π’”π’Šπ’β‘π’™ )𝒅𝒙 Let I1=∫_0^(πœ‹/2 )β–’π‘™π‘œπ‘”(𝑠𝑖𝑛π‘₯) 𝑑π‘₯ Solving 𝐈𝟏 I1=∫_0^(πœ‹/2 )β–’π‘™π‘œπ‘”(𝑠𝑖𝑛π‘₯) 𝑑π‘₯ ∴ I1=∫_𝟎^(𝝅/𝟐)β–’π’”π’Šπ’(𝝅/πŸβˆ’π’™)𝒅𝒙 I1= ∫_𝟎^(𝝅/𝟐)β–’π’π’π’ˆ(𝒄𝒐𝒔⁑𝒙 )𝒅𝒙 Adding (2) and (3) i.e. (2) + (3) 𝐈𝟏 + 𝐈𝟏 =∫_𝟎^(𝝅/𝟐)β–’γ€–π’π’π’ˆ(π’”π’Šπ’β‘π’™ )𝒅𝒙+∫_𝟎^(𝝅/𝟐)β–’π’π’π’ˆ(πœπ¨π¬β‘π’™ )𝒅𝒙〗 "Using" π’π’π’ˆβ‘π’‚ + π’π’π’ˆβ‘π’ƒ = π’π’π’ˆβ‘(𝒂.𝒃) 2I1 =∫_𝟎^(𝝅/𝟐)β–’γ€–π₯𝐨𝐠⁑[𝐬𝐒𝐧⁑〖𝒙 πœπ¨π¬β‘π’™ γ€— ] 𝒅𝒙〗 2I1 = ∫_𝟎^(𝝅/𝟐)β–’γ€–π’π’π’ˆβ‘[πŸπ’”π’Šπ’β‘γ€–π’™ 𝒄𝒐𝒔⁑𝒙 γ€—/𝟐] 𝒅𝒙〗 "Using " π’π’π’ˆ(𝒂/𝒃) = log⁑(π‘Ž) – log⁑(𝑏)) 2I1 = ∫_0^(πœ‹/2)β–’[log[2sin⁑〖π‘₯ cos⁑π‘₯ γ€— ]βˆ’π‘™π‘œπ‘”(2)]𝑑π‘₯ "Using" π’”π’Šπ’β‘πŸπ’™=2 sin⁑〖π‘₯ cos⁑π‘₯ γ€— 2I1 = ∫_0^(πœ‹/2)β–’[log[sin⁑2π‘₯ ]βˆ’π‘™π‘œπ‘”2]𝑑π‘₯ 2I1 = ∫_𝟎^(𝝅/𝟐)β–’π₯𝐨𝐠[π’”π’Šπ’β‘πŸπ’™ ]π’…π’™βˆ’βˆ«_0^(πœ‹/2)β–’log(2)𝑑π‘₯ Solving 𝐈𝟐 𝐈𝟐=∫_𝟎^(𝝅/𝟐)β–’γ€–π₯𝐨𝐠 π’”π’Šπ’β‘πŸπ’™ 𝒅𝒙〗 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▒π₯𝐨𝐠(π’”π’Šπ’β‘π’• )𝒅𝒕 Here, 𝑓(𝑑)=log⁑𝑠𝑖𝑛𝑑 𝑓(2π‘Žβˆ’π‘‘)=𝑓(2πœ‹βˆ’π‘‘)=log⁑𝑠𝑖𝑛(2πœ‹βˆ’π‘‘)=log⁑sin⁑𝑑 As, 𝒇(𝒕)=𝒇(πŸπ’‚βˆ’π’•) ∴ 𝐈𝟐 = 1/2 ∫_0^πœ‹β–’log⁑sin⁑〖𝑑 𝑑𝑑〗 =𝟏/𝟐 Γ— 𝟐∫_𝟎^(𝝅/𝟐)β–’π’π’π’ˆβ‘π’”π’Šπ’β‘γ€–π’•. 𝒅𝒕〗 =∫_0^(πœ‹/2)β–’log⁑sin⁑〖𝑑. 𝑑𝑑〗 =∫_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/𝟐)β–’γ€–πŸ.〗⁑𝒅𝒙 2I1 = I1 βˆ’ log(2) [π‘₯]_0^(πœ‹/2) 2I1βˆ’I1=βˆ’log⁑2 [πœ‹/2βˆ’0] 𝐈𝟏=βˆ’π’π’π’ˆβ‘πŸ [𝝅/𝟐] Hence, 𝐈=2 I1= 2 Γ— (βˆ’πœ‹)/2 log⁑2 =βˆ’π… π’π’π’ˆβ‘πŸ

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.