Search for: ∫0π log(1+cosx)dx is equal to ∫0π log(1+cosx)dx is equal to A0Bπ2log2C−πlog2D2πlog2 Fill Out the Form for Expert Academic Guidance!l Grade ---Class 1Class 2Class 3Class 4Class 5Class 6Class 7Class 8Class 9Class 10Class 11Class 12 Target Exam JEENEETCBSE +91 Preferred time slot for the call ---9 am10 am11 am12 pm1 pm2 pm3 pm4 pm5 pm6 pm7 pm8pm9 pm10pm Please indicate your interest Live ClassesBooksTest SeriesSelf Learning Language ---EnglishHindiMarathiTamilTeluguMalayalam Are you a Sri Chaitanya student? NoYes Verify OTP Code (required) I agree to the terms and conditions and privacy policy. Solution:Let,I=∫0π log(1+cosx)dx----iI=∫0π log{1+cos(π−x)}dx=∫0π log(1−cosx)dx[∵cos(π−x)=−cosx]… (ii) =∫0π log2sin2x2dx ∵1−cosx=2sin2x2=∫0π log2+2logsinx2dx=∫0π log2dx+2∫0π logsinx2dxIn the second. integral, put x2=t⇒dx=2dtand limits when x =0, t =0 and when x=π,t=π/2∴ I=log2[x]0π+2∫0π/2 log(sint)2dt=(log2)(π−0)+4−π2log2∵∫0π/2 logsinxdx=−π2log2=−πlog2 Related content Class 9 Science MCQ Chapter 12 Sound CBSE Class 12 English MCQ with Answers (Flamingo & Vistas) CBSE Class 7 Science MCQ Questions with Answers CBSE Class 12 Subject Wise MCQ 2024-25 ICSE Class 10 Toppers List 2024 Available Tenses Worksheet for Class 10 CBSE Class 7 Science Chapter 4 Heat MCQ Questions with Answers Nephron Diagram Class 10 CBSE Class 7 Science Chapter 2 Nutrition in Animals MCQ with Answers Rabindranath Tagore Jayanti 2024