Integration Full Chapter Explained - Integration Class 12 - Everything you need

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

Transcript

Ex 7.11, 21 The value of ∫_0^(πœ‹/2)β–’π‘™π‘œπ‘”((4 + 3 sin⁑π‘₯)/(4 + 3 cos⁑π‘₯ ))𝑑π‘₯ is 2 (B) 3/4 (C) 0 (D) βˆ’2 Let I=∫_0^(πœ‹/2)β–’π‘™π‘œπ‘”[(4 + 3 sin⁑π‘₯)/(4 + 3 π‘π‘œπ‘  π‘₯)] 𝑑π‘₯ ∴ I =∫_0^(πœ‹/2)β–’π‘™π‘œπ‘”[(4 + 3𝑠𝑖𝑛(πœ‹/2 βˆ’ π‘₯))/(4 + 3π‘π‘œπ‘ (πœ‹/2 βˆ’ π‘₯) )] 𝑑π‘₯ Using the property P4 : ∫_0^π‘Žβ–’γ€–π‘“(π‘₯)𝑑π‘₯=γ€— ∫_0^π‘Žβ–’π‘“(π‘Žβˆ’π‘₯)𝑑π‘₯ I =∫_0^(πœ‹/2)β–’π‘™π‘œπ‘”[(4 + 3π‘π‘œπ‘ π‘₯)/(4 + 3 sin⁑π‘₯ )] 𝑑π‘₯ Adding (1) and (2) i.e. (1) + (2) I +I=∫_0^(πœ‹/2)β–’π‘™π‘œπ‘”[(4 + 3 sin⁑π‘₯)/(4 + 3 cos⁑π‘₯ )] 𝑑π‘₯+∫_0^(πœ‹/2)β–’π‘™π‘œπ‘”[(4 + 3 cos⁑π‘₯)/(4 + 3 sin⁑π‘₯ )] 𝑑π‘₯ 2I = ∫_0^(πœ‹/2)β–’{π‘™π‘œπ‘”[(4 + 3 sin⁑π‘₯)/(4 + 3 cos⁑π‘₯ )]+π‘™π‘œπ‘”[(4 + 3 cos⁑π‘₯)/(4 + 3 sin⁑π‘₯ )]}𝑑π‘₯" " 2I = ∫_0^(πœ‹/2)β–’π‘™π‘œπ‘”[(4 + 3 sin⁑π‘₯)/(4 + 3 cos⁑π‘₯ )Γ—(4 + 3 cos⁑π‘₯)/(4 + 3 sin⁑π‘₯ )]𝑑π‘₯" " 2I=∫_0^(πœ‹/2)β–’γ€–log⁑1 𝑑π‘₯γ€— 2I = 0 ∴ I = 0 ∴ Option C is correct.

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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.