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: http://i.imgur.com/EwAJOOE.png

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    $\begingroup$ What area you're referring to? If you simply need to integrate this function you need to provide limits, explicitly. If you need to find some area under some segment of this curve, you need to provide a "description" of that segment, hence provide limits again, but implicitly. For example, find the area below the segment of the curve above within the half of period, or something like that. $\endgroup$ – Kaster Feb 2 '16 at 5:26
  • $\begingroup$ We need to find the area for half a cycle. Here's the question: i.imgur.com/EwAJOOE.png $\endgroup$ – Krunal Rindani Feb 2 '16 at 5:33

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 $\cos(2x)=0$


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


$x=\frac{n\pi}{2}$ 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 $1$.


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.

  • $\begingroup$ I did that. I'm facing an issue in finding the limits of integration. One of the limits is $0$ which can be clearly seen, but I can't find the second one. $\endgroup$ – Krunal Rindani Feb 2 '16 at 5:36
  • $\begingroup$ While this may be true, I think the the question is how the limits of integration change under the substitution. $\endgroup$ – pjs36 Feb 2 '16 at 5:38
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    $\begingroup$ Sin and cos have max at $1$ so area is from $0$ to $1$ $\endgroup$ – Archis Welankar Feb 2 '16 at 5:38
  • $\begingroup$ Oh. I'm familiar with changing limits under substitution. I completely forgot that $sine$ and $cosine$ had max at $1$ and min at $0$. Thanks! $\endgroup$ – Krunal Rindani Feb 2 '16 at 5:39
  • $\begingroup$ okay the limits of integration for the original equation is 0 and pi/4, since cos(2x) = 0 at pi/4. Therefore, when you use a u sub, your limits of integration change. simply plug in your upper and lower limits into u. sin(2x) evaulated at pi/4 yields 1. $\endgroup$ – David Snyder Feb 2 '16 at 5:44

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