All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 51989 by maxmathsup by imad last updated on 01/Jan/19
calculate∫π4π3sinx1+sin2xdx
Answered by peter frank last updated on 01/Jan/19
∫sinx2−cos2xdxu=cosxdu=−sinxdxdx=−dusinx∫sinx2−u2.−dusinx∫−du2−u2−122[∫du2+u−∫du2−u]−122ln[2+u2−u]+B−122ln[2+cosx2−cosx]+B
Terms of Service
Privacy Policy
Contact: info@tinkutara.com