GRE sample Math practice book problem 21, calculus

This problem is from the GRE sample test available at https://www.ets.org/s/gre/pdf/practice_book_math.pdf

Here is the problem:

$$\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\ \left ( \cos {t} + \sqrt{1+t^{2}}\sin^{3} {t}\cos^{3} {t} \right )dt$$

First I tried factoring out a cosine, but that implied integration by parts after further work. then I thought it'd be easier to attack the integral in two parts (first the cosine, then the rest). I started doing this and after a half-baked attempt to change a $\sin^{2}{t}$ into a $1-\cos^{2}{t}$, decided to use the double-angle identity:
$$\sqrt{2}+\frac{1}{8}\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\ \left (\sqrt{1+t^{2}}\sin^{3} {2t}\right )dt$$

Now I'm stuck. Maybe I was being "too clever". I cannot use the u-substitution $u = 1+t^2 \rightarrow\ du = 2t dt$ because the $2t$ is the argument of a trig function. Trig substitution seems especially hairy and likely to cause me more headaches than using an alternative strategy. I am wondering about Integration by Parts but have trouble formatting the D.I. method. Anybody have ideas on this?

I recognize that this isn't the most obtuse calculus problem but I do feel like I've spent too much time on this already and would appreciate input from anyone who can identify a 'giveaway' for an efficient strategy.

Thank you in advance for your input.

Answer

Special thanks to user Raffaele, who pointed out that odd functions integrated over [-a,a] have zero area. We have

\begin{align}
\int_{-\pi/4}^{\pi/4}\ \left ( \cos {t} + \sqrt{1+t^{2}}\sin^{3} {t}\cos^{3} {t} \right )dt & =\int_{-\pi/4}^{\pi/4}\ \cos {t}\ dt +\int_{-\pi/4}^{\pi/4}\ \sqrt{1+t^{2}}\sin^{3} {t}\cos^{3} {t} \ dt \\
& =\sqrt2 + 0\\
& = \sqrt2 \end{align}

Some further explanation:

Since $\sin{t}$ is an odd function, $\sin^3{t} $ is an odd function. Similarly, $\cos{t}$ is an even function, so $\cos^3{t}$ is an even function. Furthermore, $\sqrt{1+t^2}$ is an even function*. Now, any number of even functions times an odd function produces an odd function; thus, $\sqrt{1+t^{2}}\sin^{3} {t}\cos^{3} {t}$ is an odd function. Integrating this function over $[-\frac{\pi}{4},\frac{\pi}{4}]$ matches the format for integrating an odd function over $[-a,a]$. Therefore, $\int_{-\pi/4}^{\pi/4}\ \sqrt{1+t^{2}}\sin^{3} {t}\cos^{3} {t} \ dt=0$.

On the practice exam, the corresponding solution is in fact B), which offers $\sqrt2$.

*If it's not immediately apparent why $\sqrt{1+t^2}$ is even, recall that all even functions have the property $f(x)=f(-x)$, and notice that the square forces the argument of the root to always be positive, regardless of the sign of the input.