#### baldbrain

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...

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#### BvU

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

#### MathematicalPhysicist

Gold Member
Try the substitution $t=\sin^{2/3}s$.

#### baldbrain

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

#### baldbrain

Try the substitution $t=\sin^{2/3}s$.
Yes, that seems to do it.
$dt=(\frac{-2}{3})(\sin^{-1/3}s)(\cos{s})ds$. 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

#### baldbrain

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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