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# Crazy Tan

This is a method of integration that is very unorthodox, but it is very useful in certain situations.

# Methods

MathJax TeX Test Page The first thing you do is let $u = \tan{\frac{x}{2}}$, so $$x = 2\arctan(u) \rightarrow dx = \frac{2}{1 + u^2}du$$ By doing other trig identities, you get $$\sin{x} = \frac{2u}{1 + u^2}$$ $$\tan{x} = \frac{2u}{1 - u^2}$$ $$\text{By doing } \frac{\sin{x}}{\tan{x}} \text{ , you get } \cos{x} = \frac{1 - u^2}{1 + u^2}$$

# Example

So, basically, you just substitute trig functions into expressions in terms of u.

MathJax TeX Test Page $$\int\mathrm{\frac{1}{\sin{x} + 1}}\, \mathrm{d}x$$ $$=\int\mathrm{\frac{1}{\frac{2u}{1 + u^2} + 1}*\frac{2}{1 + u^2}}\, \mathrm{d}u$$ $$=2\int\mathrm{\frac{1}{2u + u^2 + 1}}\, \mathrm{d}u$$ $$=2\int\mathrm{\frac{1}{(u + 1)^2}}\, \mathrm{d}u$$ $$=-\frac{2}{u+1} + C$$ $$=-\frac{2}{\tan{x/2} + 1} + C$$
$$\int{\dfrac{1}{1-\sin(x)}\sqrt{\dfrac{\cos(x)}{1 + \cos(x) + \sin(x)}}dx}$$ $$\int{\dfrac{1}{1 - \frac{2u}{1 + u^2}}\sqrt{\dfrac{\frac{1-u^2}{1 + u^2}}{1 + \frac{2u}{1 + u^2} + \frac{1 - u^2}{1 + u^2}}}\dfrac{2}{1 + u^2}}du$$ $$2\int{\dfrac{1}{(u-1)^2}\sqrt{\dfrac{1-u^2}{1 + u^2 + 2u + 1 - u^2}}du}$$ $$2\int{\dfrac{1}{(u-1)^2}\sqrt{\dfrac{(1-u)(1+u)}{2(u+1)}}du}$$ $$\sqrt{2}\int{\dfrac{1}{(1-u)^2}\sqrt{1-u} du}$$ $$\sqrt{2}\int{(1-u)^{-3/2} du}$$ $$-\dfrac{2\sqrt{2}}{\sqrt{1-u}} + C$$ $$-\dfrac{2\sqrt{2}}{\sqrt{1-\tan(\frac{x}{2})}} + C$$
$$\int\mathrm{\csc(x)}\, \mathrm{d}x$$ $$\int{\dfrac{1}{\sin(x)}}\, \mathrm{d}x$$ $$\int{\dfrac{1}{\frac{2u}{1 + u^2}}\dfrac{2}{1 + u^2}}\, \mathrm{d}u$$ $$\int{\dfrac{1}{u}}\, \mathrm{d}u$$ $$=\ln(\tan(\frac{x}{2})) + C$$

David Witten