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Ex 7.11, 8 - Evaluate integral 0 -> pi log (1 + tan x) - Teachoo

Ex 7.11,8 - Chapter 7 Class 12 Integrals - Part 2
Ex 7.11,8 - Chapter 7 Class 12 Integrals - Part 3

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Ex 7.11, 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œ‹/4βˆ’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‘₯ Ex 7.11, 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œ‹/4βˆ’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‘₯ (tan(π‘Žβˆ’π‘)=tanβ‘γ€–π‘Ž βˆ’ π‘‘π‘Žπ‘› 𝑏〗/(1+γ€– tanγ€—β‘γ€–π‘Ž tan⁑𝑏 γ€— )) (As tan(πœ‹/4)=1) I=∫_0^(πœ‹/4)β–’log⁑[(1 βˆ’ tan⁑π‘₯ + 1 βˆ’ tan⁑π‘₯)/(1 + tan⁑π‘₯ )] 𝑑π‘₯ I=∫_0^(πœ‹/4)β–’log⁑[2/(1 + tan⁑π‘₯ )] 𝑑π‘₯ I=∫_0^(πœ‹/4)β–’[log⁑2 βˆ’log⁑(1+tan⁑π‘₯ ) ] 𝑑π‘₯ I=∫_0^(πœ‹/4)β–’log⁑2 𝑑π‘₯βˆ’βˆ«_0^(πœ‹/4)β–’log⁑(1+tan⁑π‘₯ ) 𝑑π‘₯ 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‘₯ ) 2I=∫_0^(πœ‹/4)β–’log⁑2 𝑑π‘₯ (Using log⁑(π‘Ž/𝑏) =logβ‘π‘Žβˆ’log⁑𝑏) …(2) 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 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.