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Evaluate the integral. sec x tan x dx from 0 to pi/4
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since y = sec(x) turns into y' = sec(x)*tan(x) this means that the integral of sec(x)*tan(x) is sec(x) + C
Now evaluate at the endpoints [sec(pi/4) + C] - [sec(0) + C] sqrt(2) - 1 So the integral of sec(x)*tan(x) from 0 to pi/4 is sqrt(2) - 1
Is that the antiderivative of sec x tan x?
yes from 0 to pi/4
Thanks.
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