Ex 7.10

Chapter 7 Class 12 Integrals
Serial order wise

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### Transcript

Ex 7.10, 21 The value of β«_0^(π/2)βπππ((4 + 3 sinβ‘π₯)/(4 + 3 cosβ‘π₯ ))ππ₯ is 2 (B) 3/4 (C) 0 (D) β2 Let I=β«_0^(π/2)βπππ[(4 + 3 sinβ‘π₯)/(4 + 3 πππ  π₯)] ππ₯ β΄ I =β«_0^(π/2)βπππ[(4 + 3π ππ(π/2 β π₯))/(4 + 3πππ (π/2 β π₯) )] ππ₯ Using the property P4 : β«_0^πβγπ(π₯)ππ₯=γ β«_0^πβπ(πβπ₯)ππ₯ I =β«_0^(π/2)βπππ[(4 + 3πππ π₯)/(4 + 3 sinβ‘π₯ )] ππ₯ Adding (1) and (2) i.e. (1) + (2) I +I=β«_0^(π/2)βπππ[(4 + 3 sinβ‘π₯)/(4 + 3 cosβ‘π₯ )] ππ₯+β«_0^(π/2)βπππ[(4 + 3 cosβ‘π₯)/(4 + 3 sinβ‘π₯ )] ππ₯ 2I = β«_0^(π/2)β{πππ[(4 + 3 sinβ‘π₯)/(4 + 3 cosβ‘π₯ )]+πππ[(4 + 3 cosβ‘π₯)/(4 + 3 sinβ‘π₯ )]}ππ₯" " 2I = β«_0^(π/2)βπππ[(4 + 3 sinβ‘π₯)/(4 + 3 cosβ‘π₯ )Γ(4 + 3 cosβ‘π₯)/(4 + 3 sinβ‘π₯ )]ππ₯" " 2I=β«_0^(π/2)βγlogβ‘1 ππ₯γ 2I = 0 β΄ I = 0 β΄ Option C is correct.