∫secxtanx dx =
secx + C
To integrate ∫secxtanx dx, we can use u-substitution. Let u = secx + tanx, then du/dx = secxtanx + sec^2x.
Now, we need to manipulate the integral so that it involves du/dx and dx only. We can rewrite secxtanx as (secx)(tanx), and use the identity tan^2x + 1 = sec^2x, to obtain:
∫(secx)(tanx) dx = ∫[(tan^2x + 1)/(secx)] dx
Substituting u = secx + tanx, we have:
∫[(tan^2x + 1)/(secx)] dx = ∫[(u^2 – 1)/u] (du/dx) dx
Notice that we can arrange the left side of the equation into (-1/u)(-u^2 + 1). Thus, we have:
∫(-1/u)(-u^2 + 1) (du/dx) dx
Integrating, we get:
∫(-1/u)(-u^2 + 1) (du/dx) dx = ∫(-1/u) du = -ln|u| + C
Substituting back in our expression for u, we have:
∫secxtanx dx = -ln|secx + tanx| + C
Therefore, the antiderivative of secxtanx is -ln|secx + tanx| + C.
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