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Integration of 1/(1+x^2) 📂Lemmas

Integration of 1/(1+x^2)

Formulas

  • Definite Integral

∫−∞∞11+x2dx=π \int_{-\infty}^{\infty} \dfrac{1}{1+x^{2}}dx = \pi

  • Indefinite Integral

∫−∞∞11+x2dx=π \int_{-\infty}^{\infty} \dfrac{1}{1+x^{2}}dx = \pi

CC is the integration constant.

Proofs

Definite Integral

Let’s substitute with x=tan⁡θx = \tan \theta. Then, the range of integration becomes ∫−∞∞→∫−π2π2\displaystyle \int_{-\infty}^{\infty} \to \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}, and since tan⁡′=sec⁡2\tan ^{\prime} = \sec^{2}, it results in dx=sec⁡2dθdx = \sec^{2} d\theta.

∫−∞∞11+x2dx=∫−π2π211+tan⁡2θsec⁡2θdθ=∫−π2π211+sin⁡2θcos⁡2θsec⁡2θdθ=∫−π2π21cos⁡2θ+sin⁡2θcos⁡2θsec⁡2θdθ=∫−π2π211cos⁡2θsec⁡2θdθ=∫−π2π2cos⁡2θsec⁡2θdθ=∫−π2π2cos⁡2θ1cos⁡2θdθ=∫−π2π2dθ=π2−(−π2)=π \begin{align*} \int_{-\infty}^{\infty} \dfrac{1}{1+x^{2}}dx &= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \dfrac{1}{1 + \tan^{2}\theta} \sec^{2} \theta d\theta \\ &= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \dfrac{1}{1 + \frac{\sin^{2} \theta}{\cos^{2} \theta}} \sec^{2} \theta d\theta \\ &= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \dfrac{1}{\frac{\cos^{2} \theta + \sin^{2} \theta}{\cos^{2} \theta}} \sec^{2} \theta d\theta \\ &= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \dfrac{1}{\frac{1}{\cos^{2} \theta}} \sec^{2} \theta d\theta \\ &= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^{2} \theta \sec^{2} \theta d\theta \\ &= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^{2} \theta \dfrac{1}{\cos^{2} \theta} d\theta \\ &= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} d\theta \\ &= \frac{\pi}{2} - (-\frac{\pi}{2}) \\ &= \pi \end{align*}

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Indefinite Integral

Similarly, by substituting with x=tan⁡θx = \tan \theta,

∫11+x2dx=∫11+tan⁡2θsec⁡2θdθ=∫cos⁡2θ1cos⁡2θdθ=∫dθ=θ+C=tan⁡−1x+C \begin{align*} \int \dfrac{1}{1+x^{2}}dx &= \int \dfrac{1}{1 + \tan^{2}\theta} \sec^{2} \theta d\theta \\ &= \int \cos^{2} \theta \dfrac{1}{\cos^{2} \theta} d\theta \\ &= \int d\theta \\ &= \theta + C \\ &= \tan^{-1} x + C \end{align*}

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