Step 1: Use the odd function shortcut.
Over a symmetric range from $-a$ to $a$, any odd function integrates to zero, because the left side cancels the right side. An odd function satisfies $f(-x) = -f(x)$. So we look for odd parts.
Step 2: Split the integrand.
Break the sum into two pieces: the part $x^3 + x\cos x + \tan^5 x$ and the lonely $+1$.
\[ I = \int_{-\pi/2}^{\pi/2}(x^3 + x\cos x + \tan^5 x)\,dx + \int_{-\pi/2}^{\pi/2} 1\,dx \]
Step 3: Check the first part is odd.
Replace $x$ with $-x$. We get $-x^3 - x\cos x - \tan^5 x$, which is the negative of the original. So this part is odd.
Step 4: Kill the odd part.
Since it is odd over a symmetric interval, its integral is zero.
\[ \int_{-\pi/2}^{\pi/2}(x^3 + x\cos x + \tan^5 x)\,dx = 0 \]
Step 5: Handle the constant.
The integral of $1$ over the interval is just the length of the interval.
\[ \int_{-\pi/2}^{\pi/2} 1\,dx = \frac{\pi}{2} - \left(-\frac{\pi}{2}\right) = \pi \]
Step 6: Add the results.
Zero plus $\pi$ gives the final value.
\[ I = 0 + \pi = \pi \]
\[ \boxed{\pi} \]