How to integrate $\int_{-\pi/4}^{\pi/4} (\cos(t) + \sqrt{1 + t^2}\cos^3(t)\sin^3(t))\;\mathrm{d}t$. How do I integrate 

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

There's no substitution that immediately jumps out at me.
 A: You don't need substitution for this one. Use the fact that $$\int_a ^b f(x)\;\mathrm{dx}=\int_a ^b f(a+b-x)\;\mathrm{dx}$$ So, replacing $t$ with $\frac {\pi}{4}+\frac {-\pi}{4}-t=-t$, we get $$I=\int_{-\pi/4}^{\pi/4}\left (\cos{(-t)} + \sqrt{1 + (-t)^2}\cos^3{(-t)}\sin^3{(-t)}\right)\;\mathrm{dt}$$ or $$I=\int_{-\pi/4}^{\pi/4} \left(\cos{(t)} - \sqrt{1 + t^2}\cos^3{(t)}\sin^3{(t)}\right)\;\mathrm{d}t$$ as $\cos (-x)=\cos x$ and $\sin (-x)=-\sin x$. We then get: $$\implies 2I=\int_{-\pi/4}^{\pi/4} 2\cos{(t)} \;\mathrm{dt}\implies I=\int_{-\pi/4}^{\pi/4} \cos{(t)} \;\mathrm{dt}$$ which means that $$I=\left(\sin \dfrac {\pi}4-\sin \left(\dfrac {-\pi}4 \right ) \right)=\sqrt 2$$
EDIT: A small proof of the fact mentioned above: Substitute $y=a+b-x$ in $\int_a ^b f(x)\;\mathrm{dx}$. The limits will now go from $b$ to $a$. The integral thus becomes $\int_b ^a f(a+b-y)(-\;\mathrm{dy})$ or $\int_a ^b f(a+b-y)\;\mathrm{dy}$.
A: Let $f(t)=\sqrt{1+t^2}\cos^3 t\sin ^3 t$ and note that $f(-t)=-f(t)$.  That is $f$ is an odd function.  Note that when we integrate any odd, integrable function around symmetric limits we have
$$\begin{align}
\int_{-a}^af(t)\,dt&=\int_{-a}^0f(t)\,dt+\int_0^af(t)\,dt\\\\
&=\int_0^af(-t)\,dt+\int_0^af(t)\,dt\\\\
&=\int_0^a(-f(t))\,dt+\int_0^af(t)\,dt\\\\
&=0
\end{align}$$
Thus, for $f(t)=\sqrt{1+t^2}\cos^3 t\sin ^3 t$ and $a=\pi/4$ we have
$$\int_{-\pi/4}^{\pi/4}\sqrt{1+t^2}\cos^3 t\sin ^3 t \,dt=0$$
Finally, 
$$\bbox[5px,border:2px solid #C0A000]{\int_{-\pi/4}^{\pi/4}(\cos t+\sqrt{1+t^2}\cos^3 t\sin ^3 t)\,dt=\sqrt{2}}$$
A: Hint: the second part is an odd function over $-\pi/4$ to $\pi/4$, which means...
