int_0^1(tan^-1x)^2/(1+x^2)dx এর মান কত?

Explanation:

Another Explanation (5): সমাধান: ধরি, \(x = \tan{\theta}\). সুতরাং, \(dx = \sec^2{\theta} d\theta = (1 + \tan^2{\theta}) d\theta = (1+x^2)d\theta\). যখন \(x = 0\), \(\theta = \tan^{-1}(0) = 0\). যখন \(x = 1\), \(\theta = \tan^{-1}(1) = \frac{\pi}{4}\). অতএব, \[ \int_0^1 \frac{(\tan^{-1}x)^2}{1+x^2} dx = \int_0^{\frac{\pi}{4}} \frac{\theta^2}{1+\tan^2\theta} (1+\tan^2\theta) d\theta = \int_0^{\frac{\pi}{4}} \theta^2 d\theta \] এখন, \(\int \theta^2 d\theta = \frac{\theta^3}{3} + C\). সুতরাং, \[ \int_0^{\frac{\pi}{4}} \theta^2 d\theta = \left[ \frac{\theta^3}{3} \right]_0^{\frac{\pi}{4}} = \frac{(\frac{\pi}{4})^3}{3} - \frac{0^3}{3} = \frac{\pi^3}{3 \cdot 4^3} = \frac{\pi^3}{3 \cdot 64} = \frac{\pi^3}{192} \] অতএব, \[ \int_0^1 \frac{(\tan^{-1}x)^2}{1+x^2} dx = \frac{\pi^3}{192} \] সুতরাং, নির্ণেয় মান \(\frac{\pi^3}{192}\). 🎉