Problems

Solution: When we apply the substitution \(x = \frac1{t}\), \(t\) runs from \(-1 \to -\infty\) as \(x\) goes from \(-1 \to 0\). Then it runs from \(\infty \to 1\) as \(x\) runs from \(0 \to 1\). So we would be able to show that: \[ \int_{-1}^{1}\frac{1}{(1+x^{2})^{2}}\,\mathrm{d}x = \int_{-1}^{-\infty}\frac{-t^{2}}{(1+t^{2})^{2}}\,\mathrm{d}t + \int_{\infty}^1 \frac{-t^{2}}{(1+t^{2})^{2}}\,\mathrm{d}t \] Let \(x = \tan u, \d x = \sec^2 u \d u\) \begin{align*} \int_{-1}^1 \frac1{(1+x^2)^2} \d x &= \int_{u = -\pi/4}^{u = \pi/4} \frac{\sec^2 u}{(1+\tan^2 u)^2} \d u \\ &= \int_{u = -\pi/4}^{u = \pi/4} \frac{1}{\sec^2 u} \d u \\ &= \int_{-\pi/4}^{\pi/4} \cos^2 u \d u \\ &= \int_{-\pi/4}^{\pi/4} \frac{1 + \cos 2 u}{2} \d u \\ &= \left [ \frac{2u + \sin 2u}{4} \right]_{-\pi/4}^{\pi/4} \\ &= \frac{\pi}{4} + \frac{1}{2} \end{align*} Let \(t = \tan u, \d t = \sec^2 u \d u\) \begin{align*} \int_{-1}^1 \frac{-t^2}{(1+t^2)^2} \d x &= \int_{u = -\pi/4}^{u = \pi/4} \frac{-\tan^2 u \sec^2 u}{(1+\tan^2 u)^2} \d u \\ &= -\int_{u = -\pi/4}^{u = \pi/4} \frac{\tan^2 u}{\sec^2 u} \d u \\ &= -\int_{-\pi/4}^{\pi/4} \sin^2 u \d u \\ &= -\int_{-\pi/4}^{\pi/4} \frac{1 - \cos 2 u}{2} \d u \\ &= -\left [ \frac{2u - \sin 2u}{4} \right]_{-\pi/4}^{\pi/4} \\ &= \frac{1}{2}-\frac{\pi}{4} \end{align*}

Solution: \begin{align*} \int_{-\frac{1}{2}\pi}^{\frac{1}{2}\pi}\frac{\cos^{2}\theta}{1-\sin\theta\sin2\alpha}\,\mathrm{d}\theta &= \int_{u = \frac12 \pi}^{u = -\frac12 \pi} \frac{\cos^2 (-u)}{1-\sin(-u) \sin 2 \alpha} -\d u \tag{\(u = -\theta\)} \\ &= \int_{\frac12 \pi}^{-\frac12 \pi} \frac{\cos^2 u}{1+\sin u \sin 2 \alpha} -\d u \\ &= \int_{-\frac12 \pi}^{\frac12 \pi} \frac{\cos^2 u}{1+\sin u \sin 2 \alpha} \d u \\ &= \int_{-\frac12 \pi}^{\frac12 \pi} \frac{\cos^2 \theta}{1+\sin \theta \sin 2 \alpha} \d \theta \\ \end{align*} Since \(\displaystyle \frac{1}{(1-a^2u^2)} = \frac12 \l \frac{1}{1+au} + \frac1{1-au} \r\) \begin{align*} \int_{-\frac12 \pi}^{\frac12 \pi} \frac{\cos^2 \theta}{1+\sin \theta \sin 2 \alpha} \d \theta &= \int_{-\frac12 \pi}^{\frac12 \pi} \frac{\cos^2 \theta}{1-\sin ^2\theta \sin^2 2 \alpha} \d \theta \\ &= \int_{-\frac12 \pi}^{\frac12 \pi} \frac{1-\sin^2 \theta}{1-\sin ^2\theta \sin^2 2 \alpha} \d \theta \\ &= \int_{-\frac12 \pi}^{\frac12 \pi} \frac{(1-\sin ^2\theta \sin^2 2 \alpha) \frac{1}{\sin^2 2\alpha} + 1 - \cosec^2 2\alpha}{1-\sin ^2\theta \sin^2 2 \alpha} \d \theta \\ &= \frac{\pi}{\sin^2 2\alpha} -\cot^2 2\alpha \int_{-\frac{\pi}2}^{\frac{\pi}2} \frac{1}{1 - \sin^2 \theta \sin^2 2 \alpha} \d \theta \\ &= \frac{\pi}{\sin^2 2\alpha} -\cot^2 2\alpha \int_{-\frac{\pi}2}^{\frac{\pi}2} \frac{1}{1 - \sin^2 \theta (1-\cos^2 2 \alpha)} \d \theta \\ &= \frac{\pi}{\sin^2 2\alpha} -\cot^2 2\alpha \int_{-\frac{\pi}2}^{\frac{\pi}2} \frac{1}{\cos^2 \theta +\sin^2 \theta \cos^2 2 \alpha} \d \theta \\ &= \frac{\pi}{\sin^2 2\alpha} -\cot^2 2\alpha \int_{-\frac{\pi}2}^{\frac{\pi}2} \frac{\sec^2 \theta}{1 +\tan^2 \theta \cos^2 2 \alpha} \d \theta \\ \end{align*} Finally, using the substitution \(u =|\cos 2 \alpha | \tan \theta, \d u = |\cos 2 \alpha |\sec^2 \theta \d \theta\) \begin{align*} \int_{-\frac{\pi}2}^{\frac{\pi}2} \frac{\sec^2 \theta}{1 +\tan^2 \theta \cos^2 2 \alpha} \d \theta &= |\sec 2\alpha|\int_{u = -\infty}^{u = \infty} \frac{1}{1 + u^2} \d u \\ &= |\sec 2 \alpha|\pi \end{align*} and so \begin{align*} I &= \frac{\pi}{\sin^2 2\alpha} -\cot^2 2\alpha |\sec 2 \alpha|\pi \\ &= \frac{\pi}{\sin^2 2\alpha} \l 1-\cos 2\alpha \r \\ &= \frac{\pi}{4\sin^2 \alpha\cos^2 \alpha} \l 2 \sin^2 \alpha \r \\ &= \frac{\pi}{2 \cos^2 \alpha} = \frac{\pi}{2} \sec^2 \alpha \end{align*} When \(\alpha\) small enough that the modulus doesn't flip the sign. When if \(\frac{1}{4}\pi<\alpha<\frac{1}{2}\pi\) we have: \begin{align*} I &= \frac{\pi}{\sin^2 2\alpha} -\cot^2 2\alpha |\sec 2 \alpha|\pi \\ &= \frac{\pi}{\sin^2 2\alpha} \l 1+\cos 2\alpha \r \\ &= \frac{\pi}{4\sin^2 \alpha\cos^2 \alpha} \l 2 \cos^2 \alpha \r \\ &= \frac{\pi}{2 \sin^2 \alpha} = \frac{\pi}{2} \cosec^2 \alpha \end{align*}