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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.

gen-ℤ ready to perish
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seeker
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    Check your limits of integration after the substitution. e.g., when $\theta=0$, $u=1+\cos 0=2$. So, the lower limit is $2$, not $1$. (I did not check your other work.) Note your answer is the negative of the answer given by the book. This would indicate that you might have the limits of integration reversed. – David Mitra May 10 '13 at 14:39
  • I see.. I noticed this but thought nothing of it, I just thought that the higher number should be placed at the top.. should it be the other way around then? 1 at the top and 2 at the bottom?.. – seeker May 10 '13 at 14:46
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    You've certainly done a thorough job of making your TeX code more complicated than it needs to be, including lots of instances of {{expression}} where {expression} suffices, and apparently you're assuming after every instance of \int you need braces around whatever follows it. I've cleaned up most of this, and set your big "display" as a "display" rather than inline. – Michael Hardy May 10 '13 at 14:47
  • @Assad Yes. The bottom limit in the original integral is $\color{red}\theta=0$. Your substitution is $u=1+\cos\color{red}\theta$. So the bottom limit becomes $u=1+\cos 0$. – David Mitra May 10 '13 at 14:48
  • @MichaelHardy I dont know how to write TeX, I use an application called MathType and just paste over whatever I have wrote. Sorry. – seeker May 10 '13 at 14:50

2 Answers2

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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.

lab bhattacharjee
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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?

Michael Hardy
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Alex
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  • I'm sorry I should have made it clearer, the question specifies I use that particularly substitution – seeker May 10 '13 at 14:45