<s>[INST] Integration by u-substitution. Evaluate $$\int_0^1 x^2 \cos\left(\frac{x^3}{3}+1\right)\cos\left(\sin\left(\frac{x^3}{3}+1\right)\right) \mathrm{d}x$$
My attempt: Let $u = \dfrac{x^3}{3}+1$ so $\mathrm{d}u = x^2\mathrm{d}x$
$$\int_0^1\cos(u)\cos(u)\mathrm{d}u$$
Then I am stuck trying to integrate $\cos^2(x)$. Any help is good. Thanks.
 [/INST] $$\int_0^1x^2\cos\left(\frac{x^3}{3}+1\right)\cos\left(\sin\left(\frac{x^3}{3}+1\right)\right)\mathrm{d}x$$
Let's let $u = \sin\left(\dfrac{x^3}{3}+1\right)$ so $\mathrm{d}u = x^2\cos\left(\dfrac{x^3}{3}+1\right)$. We also need to change the bounds of integration. So from $u = \sin\left(\dfrac{x^3}{3}+1\right) $, plugging in $\left[0,1\right]$ we get $\left[\sin(1),\ \sin\left(\dfrac43\right)\right]$ as the bounds. So now our integral reduces down to 
\begin{align*}
\int_{\sin(1)}^{\sin\left(\frac43\right)} \cos(u) \ \mathrm{d}u &= \bigg[\sin(u)\bigg]_{\sin(1)}^{\sin\left(\frac43\right)} \\
&= \sin\left(\sin\left(\frac43\right)\right) - \sin(\sin(1))
\end{align*}
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