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

Think about the parity of the second summand. You're integrating over a symmetric interval around the origin.

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

Since $\hspace{0.2cm}$$\sqrt{1+t^2}\cos^3(t)\sin^3(t)$$\hspace{0.2cm}$ is odd function and $\hspace{0.2cm}$$\cos(t)$$\hspace{0.2cm}$ is even function therefore

$\int_{-\pi/4}^{\pi/4}\sqrt{1+t^2}\cos^3(t)\sin^3(t)dt=0$

and

$\int_{-\pi/4}^{\pi/4}\cos(t)dt=2\int_{0}^{\pi/4}\cos(t)dt$

Hence

$\int_{-\pi/4}^{\pi/4}(\cos(t)+\sqrt{1+t^2}\cos^3(t)\sin^3(t))dt=2\int_{0}^{\pi/4}\cos(t)dt=2(\sin(t))|_{0}^{\pi/4}=2\times\frac{1}{\sqrt{2}}=\sqrt{2}$