All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 167024 by peter frank last updated on 04/Mar/22
∫secθtan4θdθ
Answered by cortano1 last updated on 05/Mar/22
lett=tanθI=∫t41+t2=∫t3(t)1+t2dtIBP{u=t3⇒du=3t2dtv=∫12d(1+t2)1+t2=1+t2I=t31+t2−∫3t21+t2dtI=t31+t2−3∫t(t1+t2)dtI=t31+t2−3[13t(1+t2)3−13∫(1+t2)32dt]I=t31+t2−t(1+t2)3+∫(1+t2)32dt
Terms of Service
Privacy Policy
Contact: info@tinkutara.com