Q:

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 $\int_a ^b f(a+b-y)\;\mathrm{dy}$.

