How to find the area for the curve $y=\sin^3(2x)\cos^3(2x)$?

Question

I could calculate the integration of this by substituting $u=\sin(2x)$ and could find one of the limits of integration which was $0$. However, I couldn't find second limit. The mark scheme says the limits of integration are $0$ and $1$, but I can't understand how and why. I'd really appreciate it if someone would help me out. Thanks in advance! :D

Edit: I couldn't explain the question properly so here's a screenshot of it: https://i.stack.imgur.com/xAMQ2.png

Answer 1

The limits of integration are from $x=0$ to the next value of $x$ for which $y$ is $0$, as seen in the figure.

As $$y=\sin^3(2x)\cos^3(2x)$$ $y=0$ when $\sin(2x)=0$ or $\cos(2x)=0$

Thus

$2x=n\pi$ or $2x=\frac{(2n+1)\pi}{2}$

or

$x=\frac{n\pi}{2}$ or $x=\frac{(2n+1)\pi}{4}$

The least positive value of $x$ here is $x=\frac{\pi}{4}$, which is the upper limit.

Now, if $u=sin(2x)$, then, when $x=0$, $u=\sin(0)=0$

and when $x=\frac{\pi}{4}$, $u=\sin(2\frac{\pi}{4})=\sin(\frac{\pi}{2})=1$

Hence the limits of $u$ are from $0$ to $1$.

Answer 2

use the Identity, $\cos^2(2x) = 1 - \sin^2(2x)$ Therefore, we have $\sin^3(2x)[1 - \sin^2(2x)]\cos(2x)$ Now, set $u = \sin(2x)$ and things will go smoothly from here.