<s>[INST] Definite integral containing 2 trig functions and a square root function $$\int_{-\pi/4}^{\pi/4}\bigl(\cos x+ \sqrt {1+x^2}\sin^3x \cos^3x \bigr)\, dx $$
This question is from a math GRE practice test
I've tried to solve this integral for 2 days... starting to think it is a typo
The answer is $\sqrt2$   but I need to know how to solve with steps
Some useful trig identities and derivatives:
$$\frac{d}{dx} \sin^2x = 2\cos x \sin x$$
$$\frac{d}{dx} \tan^{-1}x = \frac{1}{1+x^2}$$
$$\sin^2x = \frac{1}{2} - \frac{1}{2} \cos 2x$$
$$\cos^2x = \frac{1}{2} + \frac{1}{2} \cos 2x$$
$$\sin 2x = 2\cos x \sin x$$
 [/INST] Let $$I=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\left(\cos(x)+\sqrt{1+x^2}\sin^3(x)\cos^3(x)\right)dx$$ $$=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\cos(x)dx+\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\left(\sqrt{1+x^2}\sin^3(x)\cos^3(x)\right)dx$$
For the second integral, notice $$f(x)=\sqrt{1+x^2}\sin^3(x)\cos^3(x)$$ is an odd function since $$f(-x)=\sqrt{1+(-x)^2}\sin^3(-x)\cos^3(-x)=-\left(f(x)\right)$$
So recall that $$\int_{-b}^bf(x)dx=0$$ if $f$ is odd and the integral exists over the interval. Conclude that $\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\sqrt{1+x^2}\sin^3(x)\cos^3(x)dx=0.$
Then $$I=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\cos(x)dx+0=2\int_0^{\frac{\pi}{4}}\cos(x)dx=2\sin(x)\bigg|^{\frac{\pi}{4}}_0=2\sin\left(\frac{\pi}{4}\right)-2\sin(0)=\sqrt2$$
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