# Show that $dx = \frac{2}{1 + u^2} du$ where $u = {\tan(\frac{x}{2})}$

Hello everyone I have been trying to show that

$dx = \frac{2}{1 + u^2} du$ where $u = {\tan(\frac{x}{2})}$

but I keep ending up with something like this: $2d{\sin(\frac{x}{2})}\cos(\frac{x}{2})$ = $d\sin(\frac{2x}{x})$ = $d\sin x$

is there a mistake in the question or am I doing something wrong? thanks for the help!

• Hint: What is $d(\arctan x)$? – user122283 Feb 3 '14 at 21:32
• @Alex Have you established expression for $sinx$ or $cosx$ in terms of rational expressions $u$? Because you can derive those to find your answer. – imranfat Feb 3 '14 at 21:35

We have $$du=\left(\tan\left( \frac x 2\right)\right)'dx=\frac 1 2\left(1+\tan^2\left( \frac x 2\right)\right)dx=\frac 1 2\left(1+u^2\right)dx$$ and deduce the desired result.

• thank you! but why are we allowed the take the derivative of $\left(\tan\left( \frac x 2\right)\right)'$ – Alex Chavez Feb 3 '14 at 21:53
• We have $u=\tan\left( \frac x 2\right)$ so what's $\frac{du}{dx}$? – user63181 Feb 3 '14 at 21:58

Hint:

$$du = d(\tan(x/2)) = \left(1+\tan^2\frac{x}{2}\right)\frac{1}{2}dx,$$

so remembering that $\tan{x/2} = u$...

Cheers!

Hint: Note that $$u=\cfrac{\sin\frac x2}{\cos\frac x2}.$$ Take the derivative with respect to $x,$ using quotient and chain rules. Then, try to rewrite the result in terms of $u,$ rather than sine and cosine.