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Can you help me to show that

$$\int^{\pi/2}_{0}{d\theta\over (1-m^2\cos^2\theta)^2} \approx {(2-m^2)\pi\over4(1-m^2)^{3/2}}$$

to first order, such that $0 \lt m \lt 1$

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What do you mean, "to first order"? Do you mean, to $O(m^2)$ on either side? If so, then the LHS becomes

$$\int_{0}^{\frac{\pi}{2}} d{\theta} (1+2 m^2 \cos^2 {\theta})$$ which evaluates to $\frac{\pi}{2} (1+m^2)$. The RHS takes precisely this value upon a Taylor expansion to $O(m^2)$.

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The exercise is the following: $f(m,\epsilon)=\int_{0}^{\pi/2}{\sin^2\theta d\theta \over (1-m^2\cos^2\theta)^2sin^2\theta + \epsilon^2}$. I have split the integral into two parts at $\delta$ where $0 \lt \epsilon \ll \delta \ll 1$. I am trying to evaluate the global contribution where $\epsilon$ is negligible with error $O(\epsilon ^2)$. Does this help? – adamG Dec 19 '12 at 12:51
Sort of. Still, I think you are not formulating the problem correctly, I think. It seems to me that you are trying to find a simple, analytical expression for the original integral for all values of m that works within some error bound. Adding the epsilon in the denominator isn't going to help in this case. What will help is an expansion about m=0, and an expansion about 1/m = 0, and then a combination of the two. – Ron Gordon Dec 19 '12 at 14:58

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