# Ex 7.11, 10

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Ex 7.11, 10 By using the properties of definite integrals, evaluate the integrals : 0 𝜋2 2 log sin𝑥 − log sin2𝑥 𝑑𝑥 Let I1= 0 𝜋2 2 log sin𝑥 − log sin2𝑥 𝑑𝑥 I1= 0 𝜋2 2 log sin𝑥−𝑙𝑜𝑔 2 sin𝑥 cos𝑥 𝑑𝑥 I1= 0 𝜋2 2 log sin𝑥− log2− log sin𝑥− log cos𝑥 𝑑𝑥 I1= 0 𝜋2 log sin𝑥−𝑙𝑜𝑔2− log cos𝑥 𝑑𝑥 I1= 0 𝜋2 log sin𝑥 𝑑𝑥− 0 𝜋2 log2𝑑𝑥− 0 𝜋2 log cos𝑥 𝑑𝑥 Solving I2 I2= 0 𝜋2 log cos𝑥 𝑑𝑥 ∴ I2= 0 𝜋2 log𝑐𝑜𝑠 𝜋2−𝑥𝑑𝑥 I2= 0 𝜋2 log sin𝑥 𝑑𝑥 Put the value of I2 in (1) i.e. I1 ∴ I1= 0 𝜋2 log sin𝑥 𝑑𝑥− 0 𝜋2 log 2𝑑𝑥− 0 𝜋2 log cos𝑥 𝑑𝑥 I1= 0 𝜋2 log sin𝑥 𝑑𝑥− 0 𝜋2 log 2𝑑𝑥− 0 𝜋2 log sin𝑥 𝑑𝑥 I1= – 0 𝜋2 log 2𝑑𝑥 I1= – log 2 0 𝜋2𝑑𝑥 I1= – log 2 𝑥0 𝜋2 I1= – log 2 𝜋2−0 I1= log 2−1 𝜋2 I1= 𝜋2 log 12

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