Search for: ∫0π x1+sinxdx is equal to ∫0π x1+sinxdx is equal to AπBπ2C2πDπ8 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 x1+sinxdx----iI=∫0π π−x1+sin(π−x)dxthen, I=∫0π π−x1+sinxdx ∵∫0a f(x)dx=∫0a f(a−x)dx [∵sin(π−x)=sinx]… (ii) On adding Eqs. (i) and (ii), we get2I=∫0π π(1+sinx)dx=π∫0π 1(1+sinx)dx=π∫0π 1−sinx(1+sinx)(1−sinx)dx[multiply numerator and denominator by (1 - sin x)] ⇒ 2I=π∫0π 1−sinx1−sin2xdx =π∫0π 1cos2xdx−π∫0π sinxcos2xdx⇒ 2I=π∫0π sec2xdx−π∫0π secx⋅tanxdx⇒ 2I=π[tanx−secx]0π⇒ 2I=π[tanπ−secπ−(tan0−sec0)]⇒ 2I=π[0+1−0+1]⇒2I=2π⇒I=π 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