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

Question

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 $0 \lt m \lt 1$

Answer 1

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 $O(m^2)$.

Answer 2

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\begin{align} &\bbox[5px,#ffd]{\int_{0}^{\pi/2}{\dd\theta \over \bracks{1 - m^{2}\cos^{2}\pars{\theta}}^{2}}} \\[5mm] = &\ \int_{0}^{\pi/2}{\sec^{2}\pars{\theta} \over \bracks{\sec^{2}\pars{\theta} - m^{2}}^{2}}\, \sec^{2}\pars{\theta}\dd\theta \\[5mm] = &\ \left.\int_{0}^{\infty}{t^{2} + 1 \over \pars{t^{2} + \epsilon}^{2}}\,\dd t \,\right\vert_{\color{red}{\ {\large\epsilon}\ =\ 1 -\ m^{2}\,\, >\ 0}} \\[5mm] = &\ \pars{1 - \epsilon}\int_{0}^{\infty}{\dd t \over \pars{t^{2} + \epsilon}^{2}} + \int_{0}^{\infty}{\dd t \over t^{2} + \epsilon} \\[5mm] = & \bracks{\pars{\epsilon - 1}\partiald{}{\epsilon} + 1} \int_{0}^{\infty}{\dd t \over t^{2} + \epsilon} \\[5mm] = &\ \bracks{\pars{\epsilon - 1}\partiald{}{\epsilon} + 1} {\pi \over 2}\,\epsilon^{-1/2} \\[5mm] = &\ {\pi \over 2}\bracks{% \pars{\epsilon - 1}\pars{-\,{1 \over 2}\,\epsilon^{-3/2}} + \epsilon^{-1/2}} \\[5mm] = & {\pi \over 4}\pars{\epsilon^{-1/2} + \epsilon^{-3/2}} = {\pi \over 4}\,{\epsilon + 1 \over \epsilon^{3/2}} \\[5mm] = &\ \bbx{{\pi \over 4}\,{2 - m^{2} \over \pars{1 - m^{2}}^{3/2}}} \\[1mm] &\ \mbox{which is an}\ \underline{exact}\ \mbox{result !!!.} \\ & \end{align}

Answer 3

Note that

$$\frac d{dx}\left(\frac{m^2\sin 2\theta }{1-m^2\cos^2\theta} \right) = -\frac{2(2-m^2)}{1-m^2\cos^2\theta} + \frac{4(1-m^2)}{(1-m^2\cos^2\theta)^2} $$

Integrate both sides

$$0= -\int_0^{\pi/2} \frac{2(2-m^2)}{1-m^2\cos^2\theta}d\theta+ \int_0^{\pi/2}\frac{4(1-m^2)}{(1-m^2\cos^2\theta)^2}d\theta $$

which leads to

\begin{align} \int_0^{\pi/2}\frac{d\theta}{(1-m^2\cos^2\theta)^2} &=\frac{2-m^2}{2(1-m^2)} \int_0^{\pi/2} \frac{d\theta}{1-m^2\cos^2\theta}\\ &=\frac{2-m^2}{2(1-m^2)} \int_0^{\pi/2} \frac{d(\tan \theta)}{\tan^2\theta +1-m^2}\\ &= {\pi(2-m^2)\over4(1-m^2)^{3/2}} \end{align}