Β  Ex 7.10, 8 - Evaluate integral 0 -> pi log (1 + tan x) - Teachoo - Ex 7.10 part 2 - Ex 7.10,8 - Ex 7.10 - Serial order wise - Chapter 7 Class 12 Integrals part 3 - Ex 7.10,8 - Ex 7.10 - Serial order wise - Chapter 7 Class 12 Integrals

Share on WhatsApp

πŸŽ‰ Smart choice! You just saved 2+ minutes of ads and got straight to the good stuff. That's what being a Teachoo Black member is all about.


Transcript

Ex 7.10, 8 By using the properties of definite integrals, evaluate the integrals : ∫_0^(πœ‹/4)β–’log⁑(1+tan⁑π‘₯ ) 𝑑π‘₯ Let I=∫_0^(πœ‹/4)β–’log⁑〖 (1+tan⁑π‘₯ )γ€— 𝑑π‘₯ ∴ I=∫_0^(πœ‹/4)β–’log⁑[1+𝐭𝐚𝐧⁑(𝝅/πŸ’βˆ’π’™) ] 𝑑π‘₯ I=∫_0^(πœ‹/4)β–’log⁑[1+(tan⁑ πœ‹/4 βˆ’tan⁑π‘₯)/(1 +γ€– tan〗⁑ πœ‹/4 . tan⁑π‘₯ )] 𝑑π‘₯ I=∫_0^(πœ‹/4)β–’log⁑[1+(1 βˆ’ tan⁑π‘₯)/(1 + 1 . tan⁑π‘₯ )] 𝑑π‘₯ I=∫_0^(πœ‹/4)β–’log⁑[(1 βˆ’ tan⁑π‘₯ + 1 βˆ’ tan⁑π‘₯)/(1 + tan⁑π‘₯ )] 𝑑π‘₯ I=∫_𝟎^(𝝅/πŸ’)β–’π’π’π’ˆβ‘[𝟐/(𝟏 + 𝒕𝒂𝒏⁑𝒙 )] 𝒅𝒙 Using log⁑(π‘Ž/𝑏)=logβ‘π‘Žβˆ’log⁑𝑏 I=∫_0^(πœ‹/4)β–’[log⁑2 βˆ’log⁑(1+tan⁑π‘₯ ) ] 𝑑π‘₯ 𝐈=∫_𝟎^(𝝅/πŸ’)β–’π’π’π’ˆβ‘πŸ π’…π’™βˆ’βˆ«_𝟎^(𝝅/πŸ’)β–’π’π’π’ˆβ‘(𝟏+𝒕𝒂𝒏⁑𝒙 ) 𝒅𝒙 Adding (1) and (2) i.e. (1) + (2) I+I=∫_0^(πœ‹/4)β–’log⁑(1+tan⁑π‘₯ ) 𝑑π‘₯+∫_0^(πœ‹/4)β–’log⁑2 𝑑π‘₯βˆ’βˆ«_0^(πœ‹/4)β–’log⁑(1+tan⁑π‘₯ ) πŸπ‘°=∫_𝟎^(𝝅/πŸ’)β–’π’π’π’ˆβ‘πŸ 𝒅𝒙 2I=log⁑〖 2γ€— ∫_0^(πœ‹/4)▒𝑑π‘₯ I=log⁑〖 2γ€—/2 [π‘₯]_0^(πœ‹/4) I=log⁑2/2 [πœ‹/4 βˆ’ 0] I=log⁑2/2Γ—πœ‹/4 𝑰=𝝅/πŸ– π’π’π’ˆβ‘πŸ

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 14 years. He provides courses for Maths, Science and Computer Science at Teachoo