Given that $$y = \tan x$$ use the quotient rule to show that dy/dx = \sec^2 x --- The region enclosed by the curve $$y = \tan^2 x$$ and the horizontal line, which intersects the curve at $$x = -\frac{\pi}{4}$$ and $$x = \frac{\pi}{4}$$ is shaded in the diagram below - AQA - A-Level Maths Mechanics - Question 10 - 2021 - Paper 1

The area A of the shaded region under the curve $y = \tan^2 x$ between the limits $x = -\frac{\pi}{4}$ and $x = \frac{\pi}{4}$ is given by the integral:

$A = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \tan^2 x \, dx.$

Using the identity $\tan^2 x = \sec^2 x - 1$, the integral becomes:

$A = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} (\sec^2 x - 1) \, dx$

This can be calculated as follows:

1. Evaluate each integral separately:

   • $\int \sec^2 x \, dx = \tan x$
   • $\int 1 \, dx = x$

2. Compute their definite integrals from $-\frac{\pi}{4}$ to $\frac{\pi}{4}$:

$A = \left[ \tan x \right]_{-\frac{\pi}{4}}^{\frac{\pi}{4}} - \left[ x \right]_{-\frac{\pi}{4}}^{\frac{\pi}{4}}$

Evaluating this leads to:

$A = (1 - (-1)) - (\frac{\pi}{4} - (-\frac{\pi}{4})) = 2 - \frac{\pi}{2}$

Thus, simplifying gives:

$A = 2.$