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Ex 7.11

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Chapter 7 Class 12 Integrals

Serial order wise

Last updated at March 23, 2023 by Teachoo

Ex 7.11, 8 By using the properties of definite integrals, evaluate the integrals : ∫_0^(𝜋/4)▒log(1+tan𝑥 ) 𝑑𝑥 Let I=∫_0^(𝜋/4)▒log〖 (1+tan𝑥 )〗 𝑑𝑥 ∴ I=∫_0^(𝜋/4)▒log[1+tan(𝜋/4−𝑥) ] 𝑑𝑥 I=∫_0^(𝜋/4)▒log[1+(tan 𝜋/4 −tan𝑥)/(1 +〖 tan〗 𝜋/4 . tan𝑥 )] 𝑑𝑥 I=∫_0^(𝜋/4)▒log[1+(1 − tan𝑥)/(1 + 1 . tan𝑥 )] 𝑑𝑥 Ex 7.11, 8 By using the properties of definite integrals, evaluate the integrals : ∫_0^(𝜋/4)▒log(1+tan𝑥 ) 𝑑𝑥 Let I=∫_0^(𝜋/4)▒log〖 (1+tan𝑥 )〗 𝑑𝑥 ∴ I=∫_0^(𝜋/4)▒log[1+tan(𝜋/4−𝑥) ] 𝑑𝑥 I=∫_0^(𝜋/4)▒log[1+(tan 𝜋/4 −tan𝑥)/(1 +〖 tan〗 𝜋/4 . tan𝑥 )] 𝑑𝑥 I=∫_0^(𝜋/4)▒log[1+(1 − tan𝑥)/(1 + 1 . tan𝑥 )] 𝑑𝑥 (tan(𝑎−𝑏)=tan〖𝑎 − 𝑡𝑎𝑛 𝑏〗/(1+〖 tan〗〖𝑎 tan𝑏 〗 )) (As tan(𝜋/4)=1) I=∫_0^(𝜋/4)▒log[(1 − tan𝑥 + 1 − tan𝑥)/(1 + tan𝑥 )] 𝑑𝑥 I=∫_0^(𝜋/4)▒log[2/(1 + tan𝑥 )] 𝑑𝑥 I=∫_0^(𝜋/4)▒[log2 −log(1+tan𝑥 ) ] 𝑑𝑥 I=∫_0^(𝜋/4)▒log2 𝑑𝑥−∫_0^(𝜋/4)▒log(1+tan𝑥 ) 𝑑𝑥 Adding (1) and (2) i.e. (1) + (2) I+I=∫_0^(𝜋/4)▒log(1+tan𝑥 ) 𝑑𝑥+∫_0^(𝜋/4)▒log2 𝑑𝑥−∫_0^(𝜋/4)▒log(1+tan𝑥 ) 2I=∫_0^(𝜋/4)▒log2 𝑑𝑥 (Using log(𝑎/𝑏) =log𝑎−log𝑏) …(2) 2I=log〖 2〗 ∫_0^(𝜋/4)▒𝑑𝑥 I=log〖 2〗/2 [𝑥]_0^(𝜋/4) I=log2/2 [𝜋/4 − 0] I=log2/2×𝜋/4 𝑰=𝝅/𝟖 𝒍𝒐𝒈𝟐