Search for: If ∫e2x(cosx+7sinx)dx=e2xg(x)+C where c is a constant of integration, then g(0)+gπ2 is equal to If ∫e2x(cosx+7sinx)dx=e2xg(x)+C where c is a constant of integration, then g(0)+gπ2 is equal to 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 know that∫eaxsinbxdx=eaxa2+b2(asinbx−bcosbx)+C∫eaxcosbxdx=eaxa2+b2(acosbx+bsinbx)+C∫e2x(cosx+7sinx)dx=∫e2xcosxdx+7∫e2xsinxdx=e2x5(2cosx+sinx)+75e2x(2sinx−cosx)+C=e2x5(−5cosx+15sinx)+C∴ g(x)=15(−5cosx+15sinx)=(−cosx+3sinx)⇒ g(0)+gπ2=−1+3=2Post navigationPrevious: If ∫2×1−4xdx=Ksin−12x+C then K is equal toNext: The value of ∫1sin3xcos5xdx, is 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