All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 85718 by M±th+et£s last updated on 24/Mar/20
∫sin(x)−cos(3x)sin(x)−cos(2x)dx
Answered by MJS last updated on 24/Mar/20
∫sinx−cos3xsinx−cos2xdx==∫4cosxsin2x−cosx+sinx−2cos2x+sinx+1dx=[t=tanx2→dx=2dtt2+1]=−2∫t6+2t5−15t4+4t3+15t2+2t−1t8−2t7−4t6−6t5−10t4−6t3−4t2−2t+1dt==−2∫...(t2+1)2(t+1)2(t−2−3)(t−2+3)dt==8∫dt(t2+1)2+∫2tt2+1dt−4∫dtt2+1+23∫dt(t2+1)2−2∫dtt+1−39∫dtt−2−3+39∫dtt−2+3nowit′seasy
Commented by MJS last updated on 24/Mar/20
Iget2(5t2+6t−1)3(t2+1)(t+1)+lnt2+1(t+1)2+39lnt−2+3t−2−3witht=tanx2
Commented by M±th+et£s last updated on 24/Mar/20
thankyousir.godblessyou
Terms of Service
Privacy Policy
Contact: info@tinkutara.com