2sec^2(2x+5) dx what's the answer when dx is .3 it keeps saying...

Am I right on this? 2sec^2(2x)-8=0 2sec^2(2x)=8 divide by 2 sec^2(2x)=4 find square r...

Consider the function f(t)=2sec^2(t)–6t^2 . Let F(t) be the antiderivative of f(t) with F(0)=0 ....

f(x)= 2sec(x-3.14/4)-1

tan^2(x)=-3/2sec(x)

Prove: (tan x+sec x)^2 = 2sec^2(x)+2tan(x)secx-1

prove: (tan x + sec x)^2 = 2sec^2 x + 2tan x sec x - 1

2sec(3x-(pi/3)) Find the period.

2sec(11)sec(19)-2cot(71)

Sec^2(u/2)=(2sec(u))/(sec(u)+1)

t=2sec, s=2m,v=?

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