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Head-Scratching Integral

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Problem Statement
Evaluate $$\int{\sqrt{\frac{cosx - cos^3x} {1-cos^3x}}}\,dx$$
Relevant Equations
All the basic Integration formulae
Let I = ##\int{\sqrt{\frac{cosx - cos^3x} {1-cos^3x}}}\,dx##
I = ##\int{\sqrt{\frac{cosx(1 - cos^2x)} {1 - cos^3x}}}\,dx##
I = ##\int{\sqrt{\frac {cosx} {1 - cos^3x }}}sinx\,dx##
Substitute ##cosx = t##
Therefore, ##sinx\,dx = -dt##
So, I = ##\int{-\sqrt{\frac {t} {1 - t^3}}}\,dt##
I'm stuck here...
 

BvU

Science Advisor
Homework Helper
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the integrand does not exist as a real number in ##\left [\pi/2, 3\pi/2\right ] + 2n\pi##
 

MathematicalPhysicist

Gold Member
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Try the substitution ##t=\sin^{2/3}s##.
 
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the integrand does not exist as a real number in ##\left [\pi/2, 3\pi/2\right ] + 2n\pi##
My book says if a function is discontinuous (or doesn't exist) at a point, it's not necessary that it's anti-derivative is also discontinuous ( or doesn't exist) at that point. They've given the example of
$$\int x^{-1/3}\,dx$$ Even though the function doesn't exist at x = 0, it's anti-derivative ##\frac {x^{2/3}} {2/3} + c## is continuous at x = 0
 
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Try the substitution ##t=\sin^{2/3}s##.
Yes, that seems to do it. :smile:
##dt=(\frac{-2}{3})(\sin^{-1/3}s)(\cos{s})ds##. Also, ##\sqrt{t} = sin^{1/3}s## & ##\sqrt{1 - t^3} = cos s##
Hence, I = ##\int{\frac{-2}{3}\frac{(sin^{1/3}s)(coss)} {(coss)(sin^{1/3}s)}}\,ds##
I = ##\int{\frac{-2}{3}}\,ds## = ##\frac{-2}{3}s + c##
I = ##\frac{-2}{3}\sin^{-1}{(cos^{3/2}x)} + c##
Thanks
 
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Alternatively, the substitution ##\sqrt{1-cos^3x} = t## also works.
That yields
I = ##\frac{2}{3}\sin^{-1}{(\sqrt{1-cos^3x})} + c##
 

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