int_0^1 3/2×(tan^-1x)^2/(1+x^2)dx এর মান কত?

Explanation:

Another Explanation (5): সমাধান: ধরি, \(x = \tan(\theta)\). 🤩 তাহলে, \(dx = \sec^2(\theta) d\theta\). 😎 যখন \(x = 0\), \(\theta = 0\). যখন \(x = 1\), \(\theta = \frac{\pi}{4}\). সুতরাং, \[ \int_0^1 \frac{3}{2} \cdot \frac{(\tan^{-1}x)^2}{1+x^2} dx = \frac{3}{2} \int_0^{\frac{\pi}{4}} \frac{\theta^2}{1+\tan^2\theta} \sec^2\theta d\theta \] \[ = \frac{3}{2} \int_0^{\frac{\pi}{4}} \frac{\theta^2}{\sec^2\theta} \sec^2\theta d\theta = \frac{3}{2} \int_0^{\frac{\pi}{4}} \theta^2 d\theta \] \[ = \frac{3}{2} \left[ \frac{\theta^3}{3} \right]_0^{\frac{\pi}{4}} = \frac{3}{2} \cdot \frac{1}{3} \cdot \left( \frac{\pi}{4} \right)^3 = \frac{1}{2} \cdot \frac{\pi^3}{64} = \frac{\pi^3}{128} \] অতএব, \(\int_0^1 \frac{3}{2} \cdot \frac{(\tan^{-1}x)^2}{1+x^2} dx = \frac{\pi^3}{128}\). 🎉