<s>[INST] Evaluate $\int {\sin 2\theta \over 1 + \cos \theta} \, d\theta $, using the substitution $u = 1 + \cos \theta $ 
Evaluate $$\int_0^{\pi  \over 2} {{\sin 2\theta } \over {1 + \cos \theta }} \, d\theta$$ using the substitution $u = 1 + \cos \theta $

Using
$$\begin{align}
u &= 1 + \cos \theta \\
\frac{du}{d\theta} &= -\sin\theta \\
d\theta &= \frac{du}{-\sin\theta}\\
\end{align}$$
$$\begin{align} 
\int_0^{\pi  \over 2} {{\sin 2\theta } \over {1 + \cos \theta }}  &= \int_0^{\pi  \over 2} {{\sin 2\theta } \over {1 + \cos \theta }} {{du} \over { - \sin \theta }} \\ 
  &  = \int_0^{\pi  \over 2} {{2\sin \theta \cos \theta } \over {1 + \cos \theta }} {{du} \over { - \sin \theta }}  \\
  &  = \int_1^2 {{2(u - 1)} \over { - u}}\, du  \\ 
  &  = \int_1^2 {{2 - 2u} \over u} \,du \\
&= \int_1^2\left( {2 \over u} - 2\right) \,du  \\
  &  = \left[ 2\ln |u| - 2u \right]_1^2  \\
  &  = (2\ln 2 - 2(2)) - (2\ln1 - 2(1))  \\ 
  &  = (2\ln 2 - 4) - (2\ln 1 - 2)  \\
  &  = 2\ln 2 - 2 \\
\end{align} $$
This answer is incorrect, the answer in the book is
$$2 - 2\ln 2$$
Could someone tell me where and how I went wrong? I'm not sure, but I think it may have been when I distributed the minus sign from the denominator to the numerator.
 [/INST] Even easier $(t=\cos x$):
$$
\int_{a}^{b}\frac{\sin 2x \, dx}{1+\cos x} =-2 \int_{a}^{b} \frac{\cos x \, d \cos x}{1+\cos x}=-2\int _{\varphi(a)}^{\varphi(b)}\frac{(t+1-1) \, dt}{1+t}
$$
Can you handle from here?
</s>