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
integral of 1/ sqrt( t^2 - 9) try with substituting t = 3*sec(u) Usually trigonometric substitutions are very helpful for this kind of problems. You can get the answer as = log( 9*sqrt(t^2-9) + t))
