integral from 0 to pi/4 of (secxtanx dx)

Question

Please show how to find the antidervative of secx and tanx.

Answer 1

Well the derivative of sec(x) = sec(x) tan(x) dx, as it's one of the formulas usually memorized. So the integral of sec(x) tan(x) dx = sec(x).

But if you want a longer method, we let sec(x) = 1/cos(x) and tan(x) = sin(x) / cos(x). Substituting,
∫ sec(x) tan(x) dx
∫ (1/cos(x)) (sin(x)/cos(x)) dx
∫ (sin(x) / cos^2 (x)) dx
Let u = cos(x)
Thus du = -sin(x) dx.
∫ -du / u^2
= 1 / u
= 1 / cos(x)
= sec(x)

It's evaluated from 0 to π/4. Thus,
= sec(π/4) - sec(0)
= sqrt(2) - 1

hope this helps~ `u`

Answer 2

thank you!