Let $I=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(x^3+x\cos x+\tan^5x+1)dx$ $\Rightarrow I=I_1+\left[ x \right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}}$ $=I_1+\frac{\pi}{2}+\frac{\pi}{2}$$\Rightarrow I=I_1+\pi$where $I_1=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(x^3+x\cos x+\tan^5x)dx$Now, for $I_1$, let $f(x)=x^3+x\cos x+\tan^5x$$\therefore f(-x)=(-x)^3+(-x)\cos(-x)+\tan^5(-x)$ $=(-x^3-x\cos x-\tan^5x)=-f(x)$$\therefore f(x)$ is an odd function.Thus, $I_1=0$. Hence, $I=\pi$