12. \( \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\left(x^{3}+x^{4} \tan x\right) d x= \) (a) 0 (b) \( \frac{\pi}{2} \) (c) \( -\frac{\pi}{2} \) (d) \( \tan ^{-1}\left(\frac{1}{\sqrt{2}}\right) \) (e) \( \left(\frac{\pi}{4}\right)^{4} \tan ^{-1}\left(\frac{1}{\sqrt{2}}\right) \)

To solve the integral \( \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\left(x^{3}+x^{4} \tan x\right) d x \), we can analyze the integrand's symmetry properties. The function \( x^3 \) is an odd function, meaning it has the property that \( f(-x) = -f(x) \). Therefore, the integral of an odd function over a symmetric interval around zero (like \(-\frac{\pi}{4}\) to \(\frac{\pi}{4}\)) is zero. Conversely, \( x^4 \tan x \) is an even function, as it satisfies \( f(-x) = f(x) \). However, since the integral of the odd part \( x^3 \) contributes zero, we can conclude that the full integral evaluates to: \[ \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} (x^3 + x^4 \tan x) \, dx = 0 + \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} x^4 \tan x \, dx \] But since the integral of \( x^3 \) is zero, the resulting integral remains \( 0 \). Therefore, the answer is: (a) 0