Search for: ∫0nπ+w |sinx|dx, where n∈N and 0≤w<π is equal to∫0nπ+w |sinx|dx, where n∈N and 0≤w<π is equal toA(2n + 1) + sin wB2n + cos wC(2n + 1) - cos wDNone of these Register to Get Free Mock Test and Study Material +91 Verify OTP Code (required) I agree to the terms and conditions and privacy policy. Solution: I=∫0nπ+w |sinx|dx =∫0w |sinx|dx+∫wnπ+w |sinx|dx=I1+I2I1=∫0w |sinx|dx=∫0w sinxdx =−[cosx]0w=−cosw+1=1−coswI2=∫wnπ+w |sinx|dx=n∫0π |(sinx)|dx =n∫0π sinxdx=n[−cosx]0π=2nso, I=1−cosw+2n=(2n+1)−coswPost navigationPrevious: ∫10π+π610π+π3 (sinx+cosx)dx is equal toNext: ∫−3/210 {2x}dx where {.} denotes the fractional part of x, is equal toRelated content JEE Main 2023 Question Papers with Solutions JEE Main 2024 Syllabus Best Books for JEE Main 2024 JEE Advanced 2024: Exam date, Syllabus, Eligibility Criteria JEE Main 2024: Exam dates, Syllabus, Eligibility Criteria JEE 2024: Exam Date, Syllabus, Eligibility Criteria NCERT Solutions For Class 6 Maths Data Handling Exercise 9.3 JEE Crash Course – JEE Crash Course 2023 NEET Crash Course – NEET Crash Course 2023 JEE Advanced Crash Course – JEE Advanced Crash Course 2023