Let $f(x) = 7\tan^{8}x + 7\tan^{6}x - 3\tan^{4}x - 3\tan^{2}x$ for all $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. Then the correct expression(s) is(are)

Let f(x) = 7tan⁸x + 7tan⁶x - 3tan⁴x - 3tan²x for all x ∈ (-π/2, π/2). Then the correct expression(s) is(are)

$f(x) = 7\tan^{8}x + 7\tan^{6}x - 3\tan^{4}x - 3\tan^{2}x$

$x\in\left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right)$

$(1),(3):\ \int_{0}^{\pi/4} x f(x)\,dx = \tfrac{1}{12}\ \text{or}\ \tfrac{1}{6}$

$(2),(4):\ \int_{0}^{\pi/4} f(x)\,dx = 0\ \text{or}\ 1$

$1.\ \text{factor } f$

$2.\ \text{substitute } u=\tan x,\ \text{get } \int_0^{\pi/4} f$

$3.\ \text{parts for } \int_0^{\pi/4} x f,\ \text{reduction for } \int\tan^{n}$

$f = 7\tan^{6}x(\tan^{2}x+1) - 3\tan^{2}x(\tan^{2}x+1)$

$= (7\tan^{4}x - 3)\tan^{2}x\sec^{2}x$

$u=\tan x,\ du=\sec^{2}x\,dx;\quad x:0\to\tfrac{\pi}{4} \Rightarrow u:0\to 1$

$\int f\,dx = \int (7u^{6}-3u^{2})\,du = u^{7}-u^{3} = \tan^{7}x - \tan^{3}x = F(x)$

$F(x) = \tan^{7}x - \tan^{3}x$

$\int_{0}^{\pi/4} f\,dx = F(\tfrac{\pi}{4}) - F(0) = (1-1)-0 = 0\ \Rightarrow\ (2)$

$\int_0^{\pi/4} x f(x)\,dx = \Big[xF(x)\Big]_0^{\pi/4} - \int_0^{\pi/4} F(x)\,dx$

$\Big[xF(x)\Big]_0^{\pi/4} = \tfrac{\pi}{4}F(\tfrac{\pi}{4}) = 0$

$\int_0^{\pi/4} x f\,dx = -\int_0^{\pi/4}(\tan^{7}x-\tan^{3}x)\,dx = I_3 - I_7$

$I_n = \int_0^{\pi/4}\tan^{n}x\,dx = \frac{1}{n-1} - I_{n-2}\quad (n\ge 2)$

$I_1 = \tfrac{1}{2}\ln 2,\qquad I_3 = \tfrac{1}{2} - \tfrac{1}{2}\ln 2$

$I_7 = \tfrac{5}{12} - \tfrac{1}{2}\ln 2$

$I_3 - I_7 = \left(\tfrac{1}{2}-\tfrac{1}{2}\ln 2\right) - \left(\tfrac{5}{12}-\tfrac{1}{2}\ln 2\right) = \frac{6-5}{12} = \frac{1}{12}\ \Rightarrow\ (1)$

$\boxed{\text{Correct: (1) and (2)}}$