Homework SourseIntegral 4 to 6 1square root t2 9 dtThese are much harder th
Integral 4 to 6 ..1/square root t^2 -9 dt....These are much harder than the earlier u substitutions.. So my question is where do i begin/what do I need to look for??I can see that it is like a inverse cosh..So where do I go after kinda recognizing that...Explain to me how to go about the more difficult u-subs like these..Thank You
Solution
substitute t=3secx dt=3secxtanx so we get integral 3secxtanx dx/(3sqrt(secx^2-1)) =secxtanxdx/tanx =integral secx dx =log(tanx+secx) substituting back we get log(t/3+sqrt(t^2/9-1)) from 4 to 6 we get log(2+sqrt(4-1))-log(4/3+sqrt(16/9-1))
