The expression
    - x(t) - 1/6*x(t)^3 + 1/120*x(t)^5
looks very much like the  first few terms of the series expansion of the sine function but the signs are wrong.  I am guessing that it is meant to be
    - x(t) + 1/6*x(t)^3 - 1/120*x(t)^5

Same goes for the y(t).

How would you solve this problem with paper and pencil?  If you present an outline that shows that you understand the math, then I will be happy to show you how to code that math into Maple.

There are very many ways of doing that, some more sophisticated that others, and it's not all about knowing how to use Maple; tt also depends on what sort of mathematics you know.

Can you describe how you would construct the triangle if you were doing this a hundred years ago, before there were computers? Your (detailed) answer to that question will help us pick and present to you the solution that.best matches your background.

@mehdibgh In my original response to your question, I showed how to express the curve in the form x = my_x(y), where my_x(y) is calculated numerically.

Let's say you want to calculate the area bounded between your curve and the line x=0.5. We do:

# find limits of integration
a := fsolve(my_x(y) = 0.5, y = 0 .. 0.2);
b := fsolve(my_x(y) = 0.5, y = 1.6 .. 2.0);
area := int(0.5  - my_x(y), y = a .. b, numeric);

The answer is area = 0.6770481296.  The centroid can be calculated in the same way.

@mehdibgh Integration can be done easily in the case of the specific functions that you have provided.

Download mw.mw

@Navid555 I wrote "if u is zero, then its derivative is automatically zero, so your second condition is redundant and should be removed".  You seem to disagree.

I have nothing more to say.

I am sure that you know that if a function is identically zero on an interval, then its derivative is also identically zero on that interval.

Your initial conditions u(x, 0) = 0, D[1](u)(x, 0) = 0 say that u and its derivative are zero on the interval (0,4). But if u is zero, then its derivative is automatically zero, so your second condition is redundant and should be removed.

After removing D[1](u)(x, 0) = 0, your initial boundary value problem will have incomplete initial conditions. That's why Maple is unable to produce a solution.

Go back to the application that gives rise to the PDE and IBC and see what the missing condition is.

@acer That's an excellent approach, and I particularly like it since it is set up to be quite straightforward and transparent, though nontrivial.  I am going to be busy at work all day, but I will take a closer look tomorrow.  I am thinking of adding a to_sine::truefalse option which gives preference to expressing the result in terms of sines rather than cosines whenever possible.  For instance, 1 - cos(x)^2  will become sin(x)^2.

@acer That's an incredibly complex code which does incredibly complex things!  It does the expected:

> 2*sin(x)*cos(x);
                          2 sin(x) cos(x)
> H(%);
                               sin(2 x)

and also the unexpected:

 > 4*sin(x)^2*cos(x);
                                       2
                               4 sin(x)  cos(x)

> H(%);
                               cos(x) - cos(3 x)

I would have expected 2*sin(2*x)*sin(x) as the answer but the combine step in the definition of H intervenes and, for better or worse, takes it one step further.

@Carl Love Your combine/half_angle code is very clever and it rightly belongs to Maple's default collection of convert(...) functions.  Until then, I will keep it as an add-on in my initialization file.

As to the double_angle conversion, applying combine is not ideal. For instance

expr := 4*sin(x)*cos(x)*sin(y)*cos(y);
                     expr := 4 sin(x) cos(x) sin(y) cos(y)
combine(expr);
                   1/2 cos(-2 y + 2 x) - 1/2 cos(2 y + 2 x)

I would have preferred the answer of sin(2*x)*sin(2*y). I don't know how to construct an extension to convert that will do that.

@Carl Love Thank you very much for that example. Using subsindets is new to me and I found your example very instructive.

Here is a generalized version of your construction. I intend to place in my Maple initialization file.

Download convert-to-half-angle.mw

I also played around with the idea of making something similar for the double-angle identities:

sin(x)*cos(x) = sin(2*x) / 2;
cos(x)^2 = (1 + cos(2*x)) / 2;
sin(x)^2 = (1 - cos(2*x)) / 2;

but did not get very far.  If you have suggestions about how to go about this, I will be eager to hear them.

@acer Thanks for the explanation.  I see the fault of my original approach and I sorry for having called it a "bug".

@acer Thanks for your detailed explanation. It's clear to me now why my approach wasn't working. It's the map[2] command that I needed. Thanks again.

@Kitonum The others have pointed out what I was doing wrong and how to fix it.  That said, I very much like your approach because that we we may do:

subsindets(sin(a+b), sin(anything),t->2*sin(op(1,t)/2)*cos(op(1,t)/2));

and obtain 2 sin(a/2 + b/2) cos(a/2 + b/2)..

That brings up a question:  Is is possible to apply two transformation rules in a single call to subsindets?  Specifically, can we call subsindents just once in order to apply the two rules defined in half_angle_rule in my original post to the expression sin(x)+cos(x)?

@Carl Love Thanks for observation. Now I see what I was doing wrong.