Integrals.

Note that

$\begin{aligned} \dfrac d{dx} \tan^{\sec x} x & = \dfrac d{dx} e^{\sec x\ln \tan x} \\ & = \left(\sec x \tan x \ln (\tan x) + \sec x \cdot \frac 1{\tan x} \cdot \sec^2 x \right) e^{\sec x\ln \tan x} \\ & = \tan^{\sec x} x \left(\ln \left(\tan^{\sec x \tan x} x\right) + \frac {\sec^3 x}{\tan x} \right) \end{aligned}$

Therefore,

$\begin{aligned} \int_0^\frac \pi 4 \tan^{\sec x} x \left(\ln \left(\tan^{\sec x \tan x} x\right) + \frac {\sec^3 x}{\tan x} \right) dx & = \tan^{\sec x} x \bigg|_0^\frac \pi 4 = 1^{\sqrt 2} - 0^1 = \boxed{1} \end{aligned}$

By inspection our integral is (tanx)^(secx)+C what motivates this is when u take the derivative of exponential you have y//y=alnq+c type form so y is exponentail and part of final derivative