Computing trig integrals $$\int^{1}_{-1}(4x+1)\sin(x^2+x)\cos(x^2)dx$$I tried using substitution, however it made it seem more challenging than it had to be.  Do we first rearrange the equation, perhaps $\sin(x^2)+\sin(x)$? 
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$$\int^{1}_{-1}(4x+1)\frac{1}{2}[\sin(2x^2+x)+\sin(x)]$$ Now with $u=2x^2+x,du=4x+1$ there must be an easy way.  Does it not further compute to $$\frac{1}{2}\int^{3}_{1}\sin(u)+sin(?)$$
Or expand like suggest below,then integrate: $$\frac{1}{2}\int^{1}_{-1}[(4x)\sin(2x^2+x)+\sin(2x^2+x)+(4x)\sin(x)+sin(x)]$$
 A: As complicated the integral is, I believe the trick here is to realize that the integral of f(x), where f(x) is odd, is:
$\int_{-a}^{a} f(x) dx =0$
From the original problem we can conclude the following:
$\int_{-1}^1 (4x-1)sin(x^2+x)cos(x^2)dx$
$= \int_{-1}^1 (4x-1) ( sin(x^2)cos(x)+cos(x^2)sin(x) ) cos(x^2) dx$
$= \int_{-1}^1 ( 4xcos(x^2)sin(x^2)cos(x) + 4xcos^2(x^2)sinx -cos(x^2)sin(x^2)cosx - cos^2(x^2)sinx ) dx$
Let 
$g(x)= 4xcos(x^2)sin(x^2)cos(x)$ , and note how it is an odd function.
$m(x)= 4xcos^2(x^2)sinx$, and note how it is an even function.
$n(x)= cos(x^2)sin(x^2)cosx$, and note how it is an even function.
$p(x)= cos^2(x^2)sinx$, and note how it is an odd function.
So the integral now becomes: 
$ \int_{-1}^1 g(x) dx + \int_{-1}^1 m(x) dx - \int_{-1}^1 n(x) dx - \int_{-1}^1 p(x) dx$
$\to \int_{-1}^1 m(x) dx - \int_{-1}^1 n(x) dx$
I get stuck here, maybe someone else can jump in. I hope this has helped a bit.
A: Since
$$\sin(x^2+x)\cos(x^2) = \frac{1}{2}\left(\sin(x)+\sin(2x^2+x)\right)\tag{1}$$
our integral equals:
$$ \frac{1}{2}\int_{-1}^{1}(4x+1)\sin x\,dx +\frac{1}{2}\int_{-1}^{1}(4x+1)\sin(2x^2+x)\,dx\tag{2} $$
or just:
$$ 2\int_{-1}^{1}x\sin x\,dx -\left.\cos(2x^2+x)\right|_{-1}^{1}.\tag{3}$$
A: prove that the integrand is equals to $$\frac{1}{2} \left(4 x \sin \left(2 x^2+x\right)-\sin \left(2 x^2+x\right)+4 x
   \sin (x)-\sin (x)\right)$$ and now you can integrate
