Search for: ∫02π (sinx+|sinx|)dx is equal to ∫02π (sinx+|sinx|)dx is equal to A0B4C8D1 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:∫02π (sinx+|sinx|)dx=∫0π (sinx+sinx)dx+∫π2π (sinx−sinx)dx=∫0π 2sinxdx+∫π2π 0dx=2[−cosx]0π+0=−2(cosπ−cos0)=−2(−1−1)=4 Related content Oppositional Defiant Disorders Good Friday Wishes CBSE Class 9 Physics Motion Worksheet Autism Spectrum Disorder Sine and Cosine Waves Hindu Festivals List 2024 JEE Main Eligibility Criteria 2024 Session 2 (Released), Age Limits, Qualifying Marks, and Important Factor MCQs on Plant Hormones Class 10 5 Reasons To Choose The Commerce Stream After 10th Fl Words