Period of linear vs nonlinear pendulum I think I have an idea of how to do the problem but am not 100% sure. The question is:
(a) From the equation: $$T(E)=2\sqrt\frac{L}{g} \int_0^1 \frac{du}{u^\frac{1}{2}(1-u)^{\frac{1}{2}}[1-\frac{E(1-u)}{2g}]^\frac{1}{2}}$$ where $$u=1-\frac{g}{E}(1-cos \theta )$$
show that a nonlinear pendulum has a longer period than a linearized pendulum.
(b) Show that $\frac{dT}{dE} > 0 $. Briefly describe a physical interpretation of this result.
-------------------------------------------------------------------------------------------------------------- So I'm thinking I should derive the E equation for both a linear pendulum and non-linear pendulume (i.e E=$\frac{1}{2}mv^2+\frac{1}{2}kx_o^2$. Then plug them in and solve and see which one gives the bigger $T(E)$ for part (a). Don't know if that is the correct way to do it.  Then for part (b), I just take the derivative of $T(E)$ and show that it's $>0$. But I don't know the physical interpretation of this result.
 A: Regarding (a): the period of linear pendulum is
$$2\pi\sqrt{\frac{L}{g}}$$
and $T(E)$ is greater than that because 
$$
 \int_0^1 \frac{du}{u^\frac{1}{2}(1-u)^{\frac{1}{2}}[1-\frac{E(1-u)}{2g}]^\frac{1}{2}} >  \int_0^1 \frac{du}{u^\frac{1}{2}(1-u)^{\frac{1}{2}}} =\pi
$$
Here the last integral can be "evaluated" geometrically, by recognizing it as the arclength of the semicircle with diameter $[0,1]$. Or just do trig substitution.
For part (b), differentiate with respect to $E$ under the integral sign. You don't need to evaluate the resulting integrand; the fact that it contains a positive function is enough for conclusion. Physical interpretation: the period of nonlinear pendulum increases with amplitude. (Amplitude increases with energy, of course.)  
A: Here's another way of looking at it, perhaps more physical.
First, with the usual "force" way of looking at the pendulum, we compute the restoring force in the tangential direction, $mg\sin\theta$. Letting $m=1$ we get a differential equation for $\theta$: $$\ddot{\theta} = -g\sin\theta$$ Using the approximation $\sin\theta\approx\theta$, valid for small $\theta$, we get the equation of simple harmonic motion (SHM), and a period that's independent of amplitude. Now $\sin\theta<\theta$ (except for $\theta=0$), so the actual restoring force is always a little bit less. Thus we expect it to take a little bit longer for a period than with SHM: as the pendulum bob passes the lowest point, the force trying to bring it back is a bit weaker than for the linear pendulum, so it will take longer for it to reverse direction.
Now let's compare with the derivation you describe. I would call this "energy-based". Notice first that $g(1-\cos\theta)$ is the potential energy of the pendulum:

So $u$ is the fraction of the total energy $E$ that is kinetic: $$Eu = KE$$ $$E(1-u)=PE$$
Since the kinetic energy is $\frac{1}{2}(L\dot{\theta})^2$ (setting $m=1$), we have $dt/d\theta = L/\sqrt{2KE}$, so the period is 
$$T = \int \frac{Ld\theta}{\sqrt{2Eu}} = \int \frac{Ldu}{\sqrt{2Eu}}\frac{d\theta}{du}$$
Leaving aside the constants, this is the formula you gave. To see this, note that the things inside the last two square roots of your formula are: $$1-u=\frac{g(1-\cos\theta)}{E}$$ and
$$1-\frac{E(1-u)}{2g} = \frac{1+\cos\theta}{2}$$ and so their product is $$\frac{g}{2E}(1-\cos\theta)(1+\cos\theta) = \frac{g}{2E}\sin^2\theta$$ Differentiating the formula $u=1-(g/E)(1-\cos\theta)$ gives $du = -(g/E)\sin\theta\,d\theta$, so the formulas are equivalent.
Now, key point: what approximation do we make in omitting the last factor? Essentially, $\cos\theta\approx 1$ for small $\theta$. Just like the approximation $\sin\theta\approx\theta$ used in the "force" derivation, this is a 1st order approximation. Since $\cos\theta<1$, the integrand for the actual pendulum will be larger than for the linear pendulum, and the period will be longer.
By the way, the integral for the non-linear case is an elliptic integral of the first kind. So it can't be expressed in terms of elementary functions (i.e., finite combinations of algebraic, trig, exponentials, and logs.)
