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.
Best Answer
HINT:
$$\frac{\sin2\theta}{1+\cos\theta}=\frac{2\sin\theta\cos\theta}{1+\cos\theta}=\frac{2\sin\theta(1+\cos\theta-1)}{1+\cos\theta}=2\sin\theta -\frac{2\sin\theta}{1+\cos\theta}$$
Now put $1+\cos\theta=u$ in the second integral.