Search for: The value of ∫1sin3xcos5xdx, is The value of ∫1sin3xcos5xdx, is A−2tanx+23(tanx)3/2+CB2tanx−23(tanx)3/2+CC−2tanx+23(tanx)1/2+CDnone 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:We have,I=∫1sin3xcos5xdx⇒ I=∫1sin3/2xcos52dx⇒ I=∫sec4xtan3/2xdx [Dividing Nr and Dr by cos2 x]⇒ I=∫1+tan2xtan3/2xd(tanx)⇒ I=∫(tanx)−3/2+(tanx)1/2d(tanx)⇒ I=−2(tanx)−1/2+23(tanx)3/2+C⇒ I=−2tanx+23(tanx)3/2+CPost navigationPrevious: If ∫e2x(cosx+7sinx)dx=e2xg(x)+C where c is a constant of integration, then g(0)+gπ2 is equal to Next: ∫x+sinx1−cosxdx is equal to Related content NEET Rank Assurance Program | NEET Crash Course 2023 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