Chapter 7 Class 12 Integrals
Concept wise

  Ex 7.10, 21 (MCQ) - Value of log (4 + 3 sin x / 4 + 3 cos x) dx - Ex 7.10

part 2 - Ex 7.10, 21 (MCQ) - Ex 7.10 - Serial order wise - Chapter 7 Class 12 Integrals
part 3 - Ex 7.10, 21 (MCQ) - Ex 7.10 - Serial order wise - Chapter 7 Class 12 Integrals

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Transcript

Ex 7.10, 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‘₯ 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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