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Definite Integration by properties - P4
Ex 7.10, 2
Ex 7.10, 3 Important
Example 32 Important
Ex 7.10, 4
Ex 7.10, 15
Ex 7.10, 17
Ex 7.10, 21 (MCQ) Important
Ex 7.10, 19
Ex 7.10, 12 Important
Ex 7.10, 9
Ex 7.10,7 Important
Example 30
Question 4 (MCQ) Important Deleted for CBSE Board 2024 Exams
Ex 7.10, 10 Important You are here
Ex 7.10,8 Important
Example 34 Important
Ex 7.10, 16 Important
Question 2 Important Deleted for CBSE Board 2024 Exams
Example 42 Important
Definite Integration by properties - P4
Last updated at June 13, 2023 by Teachoo
Ex 7.10, 10 By using the properties of definite integrals, evaluate the integrals : ∫_0^(𝜋/2)▒〖 (2 logsin𝑥 −logsin2𝑥 ) 〗 𝑑𝑥 Let I1=∫_0^(𝜋/2)▒〖 (2 logsin𝑥 −logsin2𝑥 ) 〗 𝑑𝑥 I1= ∫_0^(𝜋/2)▒〖 [2 logsin𝑥 −𝑙𝑜𝑔(2 sin〖𝑥 cos𝑥 〗 )] 〗 𝑑𝑥 I1= ∫_0^(𝜋/2)▒〖 [2 logsin𝑥 −log2−logsin〖𝑥−logcos𝑥 〗 ] 〗 𝑑𝑥 I1= ∫_0^(𝜋/2)▒〖 [logsin𝑥 −𝑙𝑜𝑔2−logcos𝑥 ] 〗 𝑑𝑥 I1= ∫_0^(𝜋/2)▒logsin〖𝑥 𝑑𝑥〗 −∫_0^(𝜋/2)▒〖log2𝑑𝑥−∫_0^(𝜋/2)▒logcos〖𝑥 𝑑𝑥〗 〗 Solving I2 I2=∫_0^(𝜋/2)▒logcos〖𝑥 𝑑𝑥〗 ∴ I2= ∫_0^(𝜋/2)▒log𝑐𝑜𝑠(𝜋/2−𝑥)𝑑𝑥 I2=∫_0^(𝜋/2)▒logsin〖𝑥 𝑑𝑥〗 Put the value of I2 in (1) i.e. I1 ∴ I1= ∫_0^(𝜋/2)▒logsin〖𝑥 𝑑𝑥〗 −∫_0^(𝜋/2)▒〖log 2〗𝑑𝑥 −∫_𝟎^(𝝅/𝟐)▒𝐥𝐨𝐠𝐜𝐨𝐬〖𝒙 𝒅𝒙〗 I1= ∫_0^(𝜋/2)▒logsin〖𝑥 𝑑𝑥〗 −∫_0^(𝜋/2)▒〖log 2〗𝑑𝑥 −∫_𝟎^(𝝅/𝟐)▒𝒍𝒐𝒈𝐬𝐢𝐧〖𝒙 𝒅𝒙〗 I1= – ∫_0^(𝜋/2)▒〖log 2〗𝑑𝑥 I1= – log 2∫_0^(𝜋/2)▒𝑑𝑥 I1= – log 2[𝑥]_0^(𝜋/2) I1= – log 2[𝜋/2−0] I1= – log 2×𝜋/2 I1= log〖〖 (2)〗^(−1) 〗 [𝜋/2] I1= 𝝅/𝟐 𝐥𝐨𝐠 (𝟏/𝟐)