The Student Room Group
Reply 1
Just square the expression first (as required), and you get:

INT [e^2x - 2 + e^(-2x)] dx

which integrates without too many problems.
Reply 2
You'll be having a big surprise on how to solve this as soon as you get to P5 though... (or FP1, i dunno where they've put "hyperbolic functions" those bastards :p:)
Reply 3
mockel
Just square the expression first (as required), and you get:

INT [e^2x - 2 + e^(-2x)] dx

which integrates without too many problems.


ah..how easy!
Reply 4
what about:

INT 2 / cos²(x/2) dx
Reply 5
ThugzMansion7
what about:

INT 2 / cos²(x/2) dx

just use formula INT[(secx)^2dx] = tanx

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