1 problem found
Explain why the use of the substitution \(x=\dfrac{1}{t}\) does not demonstrate that the integrals \[ \int_{-1}^{1}\frac{1}{(1+x^{2})^{2}}\,\mathrm{d}x\quad\mbox{ and }\quad\int_{-1}^{1}\frac{-t^{2}}{(1+t^{2})^{2}}\,\mathrm{d}t \] are equal. Evaluate both integrals correctly.
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*}