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I know that my reasoning is incorrect, I just don't know where I went wrong. I did discuss this with my Maths teacher, and even she could not find what I did wrong.

Let us begin by assuming a function, $f(x)$ that is continuous and has an antiderivative in the interval $[0, 2\pi]$. Let $A$ be the area under the curve for $f(x)$ in the interval $[0, 2\pi]$

$A = \displaystyle \int_{0}^{2\pi}{f(x)\space\mathrm{d}x}$

Now there must exist a function, $g(x)$ such that:

$f(x) = g(x)\cdot \cos(x)$

Substituting the value of $f(x)$:

$A = \displaystyle\int_{0}^{2\pi}{g(x)\cdot \cos(x)\space\mathrm{d}x}$

Using t substitution:
Let $t = \sin(x)$
Then: $\mathrm{d}t = \cos(x)\space\mathrm{d}x$
And: $x = \arcsin(t)$

Changing the limits:
$t = \sin(x)$
$0$ becomes $\sin(0) = 0$
$2\pi$ becomes $\sin(2\pi) = 0$

Substituting in the definite integral:

$A = \displaystyle \int_{0}^{0}{g(\arcsin(t))\space\mathrm{d}t}$

But Definite Integral where the lower and upper bounds are the same is $0$.

$A = 0$, which is not possible.

Thanks for the help.


marked as duplicate by Matthew Towers, Lord Shark the Unknown, Vinyl_cape_jawa, Hans Lundmark, Parcly Taxel Mar 16 at 15:12

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Longer Answer :

The change of variables formula(COV) does not apply in the given conditions.

Think of the change of variables formula like this : given a set whose area you cannot find, you change it in a manner such that its area does not change and it assumes some possibly nice shape (whose area you can calculate), and the COV formula gives you license to do this, but the manner in which you "change" the set cannot be arbitrary. For example, imagine you are finding the area of a rhombus. You can "rotate and straighten" it without changing the area and get a rectangle whose area you know, and the COV formula will tell you the area of the rhombus is the area of the rectangle. However, if I decide to fold the Rhombus over itself, then I get a shape of smaller area (a triangle), so clearly the COV should not apply here, because the area has changed!

So, I have got to state by COV carefully. The most important thing, as you may have noticed, is this : the change should be "invertible". That is, the map should be injective i.e. one-one and surjective i.e. onto the new shape. The map "rotating and straightening the Rhombus" is invertible because we can just "tilt the sides and rotate back" to obtain the old shape. The map "folding the Rhombus" is not injective, because when you fold you are sending two different points to the same new point (like how when you fold paper, some points come on top of each other), and so you cannot "invert" the operation, because given the triangle the inverse map to the Rhombus maps a point to more than one point, and these are not functions, as you know. So COV would not apply there.

Now, the map $\sin : [0,2\pi] \to [-1,1]$ is not injective, for example $\sin \pi = \sin 0 = 0$. Therefore, the change of variable is not injective!

Which is exactly why you face such absurdity : the formula does not apply here. What Yves said is that the statement $\sin x = t$ implies $x = \arcsin t$ is false. Indeed, this is the case.

Because, what is $\arcsin 0$, for example? Is it $0$, because $\sin 0 = 0$? Is it $\pi$, because $\sin \pi = 0$? It cannot be both, right? Suppose it is one of them, so for example if I say that only $\arcsin 0 = 0$, then the statement $\sin x = t \implies x = \arcsin t$ will be false because $\sin \pi = 0$ but $\arcsin 0 = 0 \neq \pi$!

That is why your example is failing. Graphically speaking, you have "folded" the graph of $f(x)$ over itself (as the many preimages of the sine would describe), and the folded graph will have area zero.

So, why does "breaking the integral" work?

Well, it is because $\sin$ on those blocks is injective! For example, the function $\sin x $ on $[0,\frac \pi 2]$ is injective. Similarly on $[\frac \pi 2 ,\pi]$. So here, the function $\arcsin$ will be well-defined with unique preimage on each of these blocks, so COV will apply.

You have to now think of it this way : imagine you are finding the area of a shape. Maybe this shape is too difficult to work with, so you break it into parts. If each of these parts can be transformed injectively into a nice-enough shape, then COV will apply, you can find the area of those shapes and then add them up. For example, this can be done with a disjoint union of rhombuses (rhombi?), so you don't know how to work with the union, but you look at each rhombus, rotate and tilt the sides to get a rectangle and find the area of that.

So COV may not work on the whole shape, but maybe we can break it into parts on which the COV will work, and then add up the results.

In conclusion, your confusion is with a catch in the COV, and you have to be careful in using it. Always ensure your transformation is injective!

  • $\begingroup$ I understand now. Thank you for your answer. $\endgroup$ – Kartik Soneji Mar 15 at 13:51
  • $\begingroup$ You are welcome! Also, tell you teacher, he/she will be sweating on this one. $\endgroup$ – астон вілла олоф мэллбэрг Mar 15 at 13:55
  • $\begingroup$ Definitely. She will be thrilled to have an answer. $\endgroup$ – Kartik Soneji Mar 16 at 8:27

Short answer:

$$\sin x=t$$ doesn't mean that

$$x=\arcsin t.$$

  • 1
    $\begingroup$ @yvesdaust didn't you mean to write $x=\arcsin t$? $\endgroup$ – Pink Panther Mar 15 at 12:05
  • $\begingroup$ WHAT? My whole life has been a lie!!! But seriously though, is $\arcsin$ defined as the inverse of $\sin$? Also, does that mean that I cannot use trigonometric substitutions? Because if the upper limit would have been something other than $2\pi$, like $\pi$, then the integral would be evaluated properly. $\endgroup$ – Kartik Soneji Mar 15 at 12:06
  • $\begingroup$ @KartikSoneji Yes and no, you need to be careful: $y = \arcsin x \iff x = \sin y \color{blue}{\;\mbox{and}\; -\tfrac{\pi}{2} \le y \le \tfrac{\pi}{2}}$. $\endgroup$ – StackTD Mar 15 at 12:14
  • $\begingroup$ @PinkPanther: yes, of course ;-) $\endgroup$ – Yves Daoust Mar 15 at 12:37
  • 1
    $\begingroup$ @KartikSoneji: no, your life hasn't been a lie. You've been taught that $\sin(x)=\sin(x+2k\pi)=\sin(\pi-x+2k\pi)$. The sine is not an invertible function ! You just forgot to account for that. $\endgroup$ – Yves Daoust Mar 15 at 12:39

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