##\int_a^b x^2\sin(2x)dx## by substitution

[archaic]

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!
$$\begin{align*}
\mathrm I&=\int_a^b x^2\sin2x\,dx\\
&\text{I know that }\frac{1}{2}\frac{d(-\cos2x)}{dx}=\sin2x\\
\mathrm I&= -\frac{1}{2}\int_a^b x^2\,d(\cos2x)\\
&=-\frac{1}{8}\int_a^b \arccos(\cos2x)\arccos(\cos2x)\,d(\cos2x)\\
&=-\frac{1}{8}\int_a^b \arccos^2(\cos2x)\,d(\cos2x)\text{ I let }u=\cos2x\\
&=-\frac{1}{8}\int_{\cos2a}^{\cos2b} \arccos^2(u)\,du
\end{align*}$$

 

[PeroK]

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!
$$\begin{align*}
\mathrm I&=\int_a^b x^2\sin2x\,dx\\
&\text{I know that }\frac{1}{2}\frac{d(-\cos2x)}{dx}=\sin2x\\
\mathrm I&= -\frac{1}{2}\int_a^b x^2\,d(\cos2x)\\
&=-\frac{1}{8}\int_a^b \arccos(\cos2x)\arccos(\cos2x)\,d(\cos2x)\\
&=-\frac{1}{8}\int_a^b \arccos^2(\cos2x)\,d(\cos2x)\text{ I let }u=\cos2x\\
&=-\frac{1}{8}\int_{\cos2a}^{\cos2b} \arccos^2(u)\,du
\end{align*}$$

I think it looks okay. Although the final integral looks worse than what you started with to me.

 

[archaic]

I think it looks okay. Although the final integral looks worse that what you started with to me.

Yes it also needs an integration by parts. I just wanted to make sure the manipulation is correct.