Small big problem with integrals

[MAGNIBORO]

hi, This is a bit embarrassing But I do not understand what the problem is with this change of variable.
suppose
$$\int_{0}^{\pi }sin(u)\,du = 2$$
now make the change ## sin(u)=v ## , ## du = \frac{dv}{\sqrt{1-v^2}} ##
$$\int_{0}^{0}\frac{v}{\sqrt{1-v^2}} \, dv = 0$$
other example,
$$\int_{0}^{2\pi } n\, \, 2cos(n)sin(n) \, dn = -\pi$$
make the change ## sin^{2}(n) = x ## , ## 2 cos(n)sin(n) dn = dx ##
$$\int_{0}^{0 } \arcsin(\sqrt{x})\, \, dx = 0 $$

Now, the mistake is to suppose this? :
$$\int_{a}^{a } f(x)\, \, dx = 0$$
This is like an indeterminate form?
The correct way would be Apply limits like this?
$$\lim_{\varepsilon \rightarrow 0}\, \int_{a }^{a+\varepsilon } f(x) \, dx = 0$$

thanks

 

[mfb]

Substitution doesn't work that way if the substitution is not bijective.

In the first example, you use ##\cos x = \sqrt{1-\sin^2 x}##, but that only works as long as the cosine is positive. In the middle of your integration range it changes sign.
You can split the integral in two parts (0 to pi/2 and pi/2 to pi) and substitute in each one individually, then you get a proper non-zero integral.

 

[MAGNIBORO]

Substitution doesn't work that way if the substitution is not bijective.

In the first example, you use ##\cos x = \sqrt{1-\sin^2 x}##, but that only works as long as the cosine is positive. In the middle of your integration range it changes sign.
You can split the integral in two parts (0 to pi/2 and pi/2 to pi) and substitute in each one individually, then you get a proper non-zero integral.

thanks for replying , and the second example I guess the error is in ##\arcsin(\sqrt(x))##.
But can happen the case when solving an integral Give me the result
$$ \int_{a}^{a } f(x)\, \, dx $$
Has it ever happened to you?

 

[mfb]

That will never be the result of a proper substitution unless the initial integral has the same form, or unless you integrate along a path (e. g. in the complex plane).