Integrate sqrt(2-2cosx)dx
- Thread starter dagr8est
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I don't know if this will help much, but The Integrator gives the integral of "Sqrt[2-2Cos[x]]" as being:
. . . . .\(\displaystyle \L -2\, \sqrt{2\, -\, 2\cos{(x)}}\, \cot{\left(\frac{x}{2}\right)}\)
Eliz.
. . . . .\(\displaystyle \L -2\, \sqrt{2\, -\, 2\cos{(x)}}\, \cot{\left(\frac{x}{2}\right)}\)
Eliz.
There is a special substitution you can use. It's called the Weierstrauss substitution(I hope I spelled that right).
\(\displaystyle \L\\\int\sqrt{2-2cos(x)}dx\)
Use \(\displaystyle \L\\u=tan(\frac{x}{2});\;\ cos(x)=\frac{1-x^{2}}{1+x^{2}};\;\ dx=\frac{2}{1+u^{2}}du\)
\(\displaystyle \L\\\int\sqrt{2-\frac{1-u^{2}}{1+u^{2}}}\cdot\frac{2}{1+u^{2}}du\)
Which simplifies to:
\(\displaystyle =\L\\\int\frac{4u}{(u^{2}+1)^{\frac{3}{2}}}du\)
Now, let \(\displaystyle \L\\w=u^{2}+1;\;\ dw=2udu;\;\ \frac{dw}{2}=udu\)
\(\displaystyle \L\\2\int\frac{1}{w^{\frac{3}{2}}}dw\)
\(\displaystyle =\L\\\frac{-4}{\sqrt{w}}\)
Resub:
\(\displaystyle \L\\\frac{-4}{\sqrt{u^{2}+1}}\)
Resub again:
\(\displaystyle \L\\\frac{-4}{\sqrt{tan^{2}(\frac{x}{2})+1}}+C\)
As Soroaban says, TA...DAA!!!
This works on the interval 0 to Pi.
\(\displaystyle \L\\\int\sqrt{2-2cos(x)}dx\)
Use \(\displaystyle \L\\u=tan(\frac{x}{2});\;\ cos(x)=\frac{1-x^{2}}{1+x^{2}};\;\ dx=\frac{2}{1+u^{2}}du\)
\(\displaystyle \L\\\int\sqrt{2-\frac{1-u^{2}}{1+u^{2}}}\cdot\frac{2}{1+u^{2}}du\)
Which simplifies to:
\(\displaystyle =\L\\\int\frac{4u}{(u^{2}+1)^{\frac{3}{2}}}du\)
Now, let \(\displaystyle \L\\w=u^{2}+1;\;\ dw=2udu;\;\ \frac{dw}{2}=udu\)
\(\displaystyle \L\\2\int\frac{1}{w^{\frac{3}{2}}}dw\)
\(\displaystyle =\L\\\frac{-4}{\sqrt{w}}\)
Resub:
\(\displaystyle \L\\\frac{-4}{\sqrt{u^{2}+1}}\)
Resub again:
\(\displaystyle \L\\\frac{-4}{\sqrt{tan^{2}(\frac{x}{2})+1}}+C\)
As Soroaban says, TA...DAA!!!
This works on the interval 0 to Pi.
skeeter
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I tried this ...
\(\displaystyle \L \sqrt{2 - 2cosx} = \sqrt{2}\sqrt{1 - cosx}\)
\(\displaystyle \L \sqrt{1 - cosx}*\frac{\sqrt{1 + cosx}}{\sqrt{1 + cosx}} =\)
\(\displaystyle \L \frac{\sqrt{1 - cos^2x}}{\sqrt{1 + cosx}} =\)
\(\displaystyle \L \frac{\sqrt{sin^2x}}{\sqrt{1 + cosx}} =\)
for \(\displaystyle \L sinx > 0\) ...
\(\displaystyle \L \frac{sinx}{\sqrt{1 + cosx}}\)
\(\displaystyle \L \sqrt{2} \int \frac{sinx}{\sqrt{1 + cosx}} dx\)
let \(\displaystyle \L u = 1 + cosx\)
\(\displaystyle \L du = -sinx dx\)
\(\displaystyle \L -\sqrt{2} \int \frac{-sinx}{\sqrt{1 + cosx}} dx\)
\(\displaystyle \L -\sqrt{2} \int \frac{1}{\sqrt{u}} du\)
\(\displaystyle \L -\sqrt{2}(2\sqrt{u}) + C\)
\(\displaystyle \L -2\sqrt{2}\sqrt{1 + cosx} + C\)
\(\displaystyle \L \sqrt{2 - 2cosx} = \sqrt{2}\sqrt{1 - cosx}\)
\(\displaystyle \L \sqrt{1 - cosx}*\frac{\sqrt{1 + cosx}}{\sqrt{1 + cosx}} =\)
\(\displaystyle \L \frac{\sqrt{1 - cos^2x}}{\sqrt{1 + cosx}} =\)
\(\displaystyle \L \frac{\sqrt{sin^2x}}{\sqrt{1 + cosx}} =\)
for \(\displaystyle \L sinx > 0\) ...
\(\displaystyle \L \frac{sinx}{\sqrt{1 + cosx}}\)
\(\displaystyle \L \sqrt{2} \int \frac{sinx}{\sqrt{1 + cosx}} dx\)
let \(\displaystyle \L u = 1 + cosx\)
\(\displaystyle \L du = -sinx dx\)
\(\displaystyle \L -\sqrt{2} \int \frac{-sinx}{\sqrt{1 + cosx}} dx\)
\(\displaystyle \L -\sqrt{2} \int \frac{1}{\sqrt{u}} du\)
\(\displaystyle \L -\sqrt{2}(2\sqrt{u}) + C\)
\(\displaystyle \L -2\sqrt{2}\sqrt{1 + cosx} + C\)