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  1. Chapter 7 Class 12 Integrals
  2. Serial order wise

Transcript

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
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.