<s>[INST] Integrate using Trigonometric Substitutions Evaluate the integral using trigonometric substitutions.  
$$\int{  x\over \sqrt{3-2x-x^2}} \,dx$$
I am familiar with using the right triangle diagram and theta, but I do not know which terms would go on the hypotenuse and sides in this case. If you can determine which numbers or $x$-values go on the hypotenuse, adjacent, and opposite sides, I can figure out the rest, although your final answer would help me check mine. Thanks!
 [/INST] $\int \frac{x}{\sqrt{4-(x+1)^2}}dx = \int \frac{2\sin\theta-1}{\sqrt{4-4\sin^2\theta}}(2\cos\theta)d\theta$
(using the substitution $x+1=2\sin\theta$)
$=\int\frac{2\sin\theta-1}{2\cos\theta}2\cos\theta d\theta$
$= \int (2\sin\theta-1) d\theta$
$=-2\cos\theta-\theta +C$
$=-2\left(\frac{\sqrt{3-2x-x^2}}{2}\right) - \sin^{-1}\left(\frac{x+1}{2}\right)+C$
$=-\sqrt{3-2x-x^2}- \sin^{-1}\left(\frac{x+1}{2}\right)+C$
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