Differentiating an accumulation function via FTC Part 1

By the Fundamental Theorem of Calculus Part 1, since the integrand $f(t) = t\tan(t)$ is continuous near $t = \pi/4$, we have $F'(x) = f(x) = x\tan(x)$ for all such $x$. This directly gives the derivative of the accumulation function without needing to compute the integral itself.

Substituting $x = \pi/4$ gives $F'(\pi/4) = \dfrac{\pi}{4}\tan\left(\dfrac{\pi}{4}\right)$. Since $\tan(\pi/4) = 1$, this simplifies to $F'(\pi/4) = \dfrac{\pi}{4}$.

The function substituted was the original integrand, t times tangent of t, rewritten with x in place of t. By FTC Part 1, F prime of x equals this integrand evaluated at x, namely x tangent x. No differentiation of the integral itself was needed, only replacing t with x in the integrand.