int_0^1(4(tan^-1x)^2/(1+x^2)   এর মান কত? 

Explanation:

Another Explanation (5): সমাধান: ধরি, \(x = \tan(\theta)\). সুতরাং, \(dx = \sec^2(\theta) d\theta\) এবং \(\tan^{-1}(x) = \theta\). যখন \(x = 0\), \(\theta = \tan^{-1}(0) = 0\). যখন \(x = 1\), \(\theta = \tan^{-1}(1) = \frac{\pi}{4}\). তাহলে, প্রদত্ত ইন্টিগ্রালটি হবে: \[ \int_{0}^{1} \frac{4(\tan^{-1}x)^2}{1+x^2} dx = \int_{0}^{\frac{\pi}{4}} \frac{4\theta^2}{1+\tan^2\theta} \sec^2\theta d\theta \] যেহেতু \(1 + \tan^2\theta = \sec^2\theta\), \[ \int_{0}^{\frac{\pi}{4}} \frac{4\theta^2}{\sec^2\theta} \sec^2\theta d\theta = \int_{0}^{\frac{\pi}{4}} 4\theta^2 d\theta \] এখন ইন্টিগ্রেট করি: \[ 4 \int_{0}^{\frac{\pi}{4}} \theta^2 d\theta = 4 \left[ \frac{\theta^3}{3} \right]_{0}^{\frac{\pi}{4}} \] \[ = 4 \left( \frac{(\frac{\pi}{4})^3}{3} - \frac{0^3}{3} \right) = 4 \cdot \frac{\pi^3}{3 \cdot 4^3} = 4 \cdot \frac{\pi^3}{3 \cdot 64} = \frac{\pi^3}{3 \cdot 16} = \frac{\pi^3}{48} \] অতএব, \(\int_{0}^{1} \frac{4(\tan^{-1}x)^2}{1+x^2} dx = \frac{\pi^3}{48}\). 🎉