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One of the integrals is:

$$\int \frac{\mathrm{d}x}{2+2\sin x + \cos x}\, \mathrm{d}x $$

How can there be two $\mathrm{d}x$?

MIT's Integration Bee

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13  
It's a "typo"... (one I sometimes make) – David Mitra Dec 10 '12 at 14:43
2  
@DavidMitra Well, its from MIT and I thought maybe it was some strange high level math, or it was a typo. It's is a fun list of integrals to practice my skill one. – yiyi Dec 10 '12 at 14:47
1  
It is a 'typo', definitely. – 0x2207 Dec 10 '12 at 14:58
6  
+1 for citing the source. – Noah Snyder Dec 10 '12 at 15:18
up vote 10 down vote accepted

The extra $dx$ is definitely a typo. To solve this integral we have to substitute $u=\tan\frac{x}{2}$. Then, $\sin x=\frac{2u}{1+u^2}$, $\cos x=\frac{1-u^2}{1+u^2}$ and $dx=\frac{du}{1+u^2}$. Thus, $$\int\frac{dx}{2+2\sin x+\cos x}=\int\frac{du}{2+2u^2+4u+1-u^2}=\int\frac{du}{u^2+4u+3}=\int\frac{du}{(u-1)(u-3)}=\frac{1}{2}\int\frac{du}{1-u}+\frac{1}{2}\int\frac{du}{u-3}=\frac{1}{2}\ln\left|1-u\right|+\frac{1}{2}\ln\left|u-3\right|+c=\\ \frac{1}{2}\ln\left|1-\tan\frac{ x}{2}\right|+\frac{1}{2}\ln\left|\tan\frac{ x}{2}-3\right|+c$$

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