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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

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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 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.