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.
 A: $$\int_0^1x^2\cos\left(\frac{x^3}{3}+1\right)\cos\left(\sin\left(\frac{x^3}{3}+1\right)\right)\mathrm{d}x$$
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*}
A: Let $\sin \left(\frac{x^3}{3}+1\right)=u$, then  $x^2 \cos \left(\frac{x^3}{3}+1\right)\mathrm dx=\mathrm du$,
Using all these we have, $$\int x^2 \cos\left(\frac{x^3}{3}+1\right)\cos\left(\sin\left(\frac{x^3}{3}+1\right)\right) \mathrm{d}x=\int \cos(u) \mathrm du$$
can you finish from here?
A: You are on the right track. When you make the substitution, don't forget to change the limits of integration as well. So we get
$$\int_0^1 x^2 \cos\left(\frac{x^3}{3}+1\right)\cos\left(\sin\left(\frac{x^3}{3}+1\right)\right) \mathrm{d}x = \int_1^{\frac{4}{3}} \cos(u)\cos(\sin(u)) \mathrm{d}x$$
Now, by making another substitution $v=\sin(u)$, $\mathrm{d}v=\cos(u)\mathrm{d}u$, we get
\begin{align*}
\int_1^{\frac{4}{3}} \cos(u)\cos(\sin(u)) \mathrm{d}x &= \int_{\sin{1}}^{\sin{\frac{4}{3}}}\cos v \mathrm{d}v\\
&= \sin\sin\frac{4}{3}-\sin\sin1
\end{align*}
Of course, we could have gotten this by making the substitution $v = \sin\left(\frac{x^3}{3}+ 1\right)$ in the first place.
A: Let $$u = \sin\left(\frac{x^3}{3}+1\right)$$
$$ \frac{du}{dx} = x^2\cos\left(\frac{x^3}{3}+1\right)$$
$$\int_0^1 \cos u \ du  = \left(\sin \sin\left(\frac{x^3}{3}+1\right) \right)_0^1 = \sin \sin\left(\frac{4}{3}\right) - \sin \sin\left(1\right) =0.08$$
A: $$\int_0^1 x^2 \cos\left(\frac{x^3}{3}+1\right)\cos\left(\sin\left(\frac{x^3}{3}+1\right)\right) dx$$
Let $u=\sin\left(\frac{x^3}{3}+1\right)$, then
$$du=x^2\cos\left(\frac{x^3}{3}+1\right)dx$$
So now
$$\int_{\sin 1}^{\sin\frac43} \cos u\ du=\sin\left(\sin \frac43 \right)-\sin\left(\sin 1\right)$$
