Impossible identity? $ \tan{\frac{x}{2}}$ $$\text{Let}\;\;t = \tan\left(\frac{x}{2}\right). \;\;\text{Show that}\;\dfrac{dx}{dt} = \dfrac{2}{1 + t^2}$$
I am saying that this is false because that identity is equal to $2\sec^2 x$ and that can't be equal. Also if I take the derivative of an integral I get the function so if I take the integral of a derivative I get the function also so the integral of that is $x + \sin x$ which evaluated at 0 is not equal to $\tan(x/2)$

Paul. You've asked a couple of nice questions here, if I understand you right:

1) Given that $t$ can be expressed as a function of $x$ (specifically, you asked about the case where $t = \tan(\frac{x}{2})$), how can we find $\frac{dx}{dt}$, if it exists?
2) If we know that $\tan(\frac{x}{2}) = t$, then why does $x = 2\arctan(t) + 2n\pi$? 

The answer to the first question comes from the chain rule. By the chain rule, we can write
$$\frac{dx}{dx} = \frac{dx}{dt} \frac{dt}{dx};$$
since we know that $\frac{dx}{dx} = 1$, this tells us that
$$ 1 = \frac{dx}{dt} \frac{dt}{dx} \implies \frac{dx}{dt} = \frac{1}{\frac{dt}{dx}}.$$
(This proof assumes that $x$ can be expressed as a differentiable function of $t$, but that assumption is valid here, so all is well.)
In your case, we see that 
$$ t = \tan(\frac{x}{2}) \implies \frac{dt}{dx} = \sec^2(\frac{x}{2}) \cdot \frac{1}{2} $$
by the chain rule, and so
$$\frac{dx}{dt} = \frac{1}{\frac{dt}{dx}} = \frac{2}{\sec^2(\frac{x}{2})}$$
But $\sec^2(x) = 1 + \tan^2(x)$, and so 
$$\frac{dx}{dt} = \frac{2}{\sec^2(\frac{x}{2})} = \frac{2}{1 + \tan^2(\frac{x}{2})} = \frac{2}{1 + t^2} $$
where to get the last equality I just plugged in $\tan(\frac{x}{2}) = t$.
Your second question has a fairly simple answer: the $\arctan(t)$ function as mathematicians usually define it outputs an angle $\theta$ between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$, such that $\tan(\theta)$ = $t$. We have to bear in mind that, when we take $\arctan(\tan(\frac{x}{2}))$, we may not get back $\frac{x}{2}$ - just $\theta$. However, we know for certain that $\frac{x}{2} = \theta + n\pi$ for some integer $n$ and some $\theta$ between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ (ponder this if it is not clear). This $\theta$ will be precisely the one which $\arctan$ returns for the argument $t = \tan(\frac{x}{2})$, and so this $\theta$ is $\arctan(t)$. Thus,
$$\frac{x}{2} = \arctan(t) + n\pi$$
which is the second thing you were having trouble with.
Hope this helps! 
