How to quickly solve $y=\int_{-\pi/4 }^{\pi/4 } \left[\cos x + \sqrt{1+x^2}\sin^3x\cos^3x\right]dx$?

Question

I'm currently trying to solve $$y=\int_{-\pi/4 }^{\pi/4 } \left[\cos x + \sqrt{1+x^2}\sin^3x\cos^3x\right]dx,$$ which is a GRE math subject test problem. I was able to get the answer ($\sqrt{2}$) by breaking up the integral $$\int_{-\pi/4 }^{\pi/4 } \cos x \ dx+\int_{-\pi/4 }^{\pi/4 } \sqrt{1+x^2}\sin^3x\cos^3x \ dx.$$ Then I used $$1+x^2=(z-x)^2$$ to get $$t=\frac{z^2-1}{2z}$$ as well as $$\sin x = \frac{2z}{1+z^2}, \ \ \cos x = \frac{1-z^2}{1+z^2}, \ \ dx = \frac{2 \ dz}{1+z^2}.$$ Note that $$z=\tan(x/2).$$ I then plugged these into $$\int_{-\pi/4 }^{\pi/4 }\sqrt{1+x^2}\sin^3x\cos^3x \ dx$$ to get something rather ugly \begin{align*} &\int_{-\tan{\pi/8}}^{\tan{\pi/8}}\sqrt{1+ \frac{z^2-1}{2z}^2 } \cdot\left( \frac{2z}{1+z^2}\right)^3\cdot\left(\frac{1-z^2}{1+z^2}\right)^3\cdot\left(\frac{2}{1+z^2}\right)dz \\ &= \int_{-\tan{\pi/8}}^{\tan{\pi/8}} \frac{16 z^3 (1 - z^2)^3 \sqrt{\frac{(z^2 - 1)^2}{4 z^2} + 1}}{(z^2 + 1)^7}dz. \end{align*} Since this is an odd function, the solution is $0$ and thus $$y=\int_{-\pi/4 }^{\pi/4 } \cos x \ dx + 0 =\sqrt{2}.$$ Even though I was able to get to the answer, I'd imagine on the math GRE subject test, I would not have had enough time to figure all this out. Are there any tricks for quickly solving definite integrals like this?

Answer 1

Hint. The function $$ x \mapsto \sqrt{1+x^2}\sin^3x\cos^3x, \qquad x \in \left[-\frac{\pi}4, \frac{\pi}4\right] $$ is odd as is $x \mapsto \sin (x)$.

Answer 2

If you noticed that $\sqrt{1+x^2}\sin^3 x \cos^3 x$ was an odd function, you would immediately get that $$\displaystyle\int_{-\pi/4}^{\pi/4}\sqrt{1+x^2}\sin^3 x \cos^3 x\,dx = 0,$$

and thus, the given integral simplifies to just $y = \displaystyle\int_{-\pi/4}^{\pi/4}\cos x\,dx$.