To evaluate the integral ∫(3π/4, -π/4) 7 sec(x) tan(x) dx, we can use trigonometric identities to simplify the integrand. Let me work it out.
First, let's simplify the integrand using the identity sec(x) = 1/cos(x) and tan(x) = sin(x)/cos(x):
7 sec(x) tan(x) = 7 * (sin(x)/cos(x)) * (1/cos(x)) = 7 * sin(x)/cos^2(x)
Now, we can rewrite the integral as:
∫(3π/4, -π/4) 7 * sin(x)/cos^2(x) dx
To integrate, let's use the substitution method. Let u = cos(x), then du = -sin(x) dx:
∫(3π/4, -π/4) -7 du/u^2
Now, integrate:
∫(3π/4, -π/4) -7 du/u^2 = [-7/u] evaluated from 3π/4 to -π/4
= [-7/cos(x)] from 3π/4 to -π/4
Now, plug in the limits:
= [-7/cos(-π/4)] - [-7/cos(3π/4)]
= [-7/(√2)] - [-7/0] (cos(3π/4) = 0)
The integral is divergent because the denominator becomes 0 at x = 3π/4.