Doing $\int_a^b x^2\sin(2x)dx$ by "substitution"

Question

Would this be valid manipulation for $x\in[0,\,\pi/2]$? I know that it is integrable by parts, I just want to know where did the manipulation become invalid, if it did, and why. Thank you! $$\begin{align*} \mathrm I&=\int_a^b x^2\sin2x\,dx\\ &\text{I know that }\frac{1}{2}\frac{d(-\cos2x)}{dx}=\sin2x\\ \mathrm I&= -\frac{1}{2}\int_a^b x^2\,d(\cos2x)\\ &=-\frac{1}{8}\int_a^b \arccos(\cos2x)\arccos(\cos2x)\,d(\cos2x)\\ &=-\frac{1}{8}\int_a^b \arccos^2(\cos2x)\,d(\cos2x)\text{ I let }u=\cos2x\\ &=-\frac{1}{8}\int_{\cos2a}^{\cos2b} \arccos^2(u)\,du \end{align*}$$

Answer 1

The only dodgy thing you've done is assume that $2x=\arccos(\cos(2x))$. This will only hold if $2x$ is in the range of $\arccos$: that is, if $0 \le 2x \le \pi$, or $0 \le x \le \frac{\pi}{2}$. In other words, if the interval $[a,b]$ is entirely contained in the interval $\left[0,\frac{\pi}{2}\right]$, then your manipulations are valid. Otherwise, they are not.

That being said, I don't think your manipulations are useful regardless of the interval of integration. Evaluating the integral $\int \arccos^2 u \, du$ requires you to integrate by parts twice, just as evaluating the original integral did. So you haven't really made any progress.