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


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

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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.