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

Question

Found this problem on the most recent GRE practice booklet by ets. I ended up skipping it because it looked like too much work. Going back through, I still don't know how to evaluate it.

To me, this is the obvious first step to make.

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

Having taken a few practice test, I have a gut feeling that the integral that remains will vanish, especially since the integral is over $[-\frac{\pi}{4},\frac{\pi}{4}]$, however I'm not sure how to show that this is indeed true. Is there an easy/quick way to reason through this?

Answer 1

Yes, the integrand is an odd function: $$ \sqrt{1+t^2} \cos^3(t)\sin^3(t)=-\sqrt{1+(-t)^2} \cos^3(-t)\sin^3(-t). $$

Answer 2

HINT:

$$\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \left(\cos(t)+\sqrt{1+t^2} \cos^3(t)\sin^3(t)\right) \text{d}t =$$ $$\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \cos(t)\text{d}t+\frac{1}{32}\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\left(3\sqrt{t^2+1}\sin(2t)-\sqrt{t^2+1}\sin(6t)\right)\text{d}t =$$ $$\left[\sin(t)\right]_{-\frac{\pi}{4}}^{\frac{\pi}{4}}+\frac{1}{32}\left(\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}3\sqrt{t^2+1}\sin(2t)\text{d}t-\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\sqrt{t^2+1}\sin(6t)\text{d}t\right)=$$ $$\left[\sin(t)\right]_{-\frac{\pi}{4}}^{\frac{\pi}{4}}+\frac{1}{32}\left(3\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\sqrt{t^2+1}\sin(2t)\text{d}t-\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\sqrt{t^2+1}\sin(6t)\text{d}t\right)$$