16/π^2int_0^1(tan^-1x)^2/(1+x^2)dx = কত?

Explanation:

Another Explanation (5): bài toán: \(\frac{16}{\pi^2} \int_0^1 \frac{(\tan^{-1}x)^2}{1+x^2} dx = ?\) giải: let \(u = \tan^{-1}x\), then \(du = \frac{1}{1+x^2} dx\). when \(x = 0\), \(u = \tan^{-1}(0) = 0\). when \(x = 1\), \(u = \tan^{-1}(1) = \frac{\pi}{4}\). so, the integral becomes: \(\int_0^1 \frac{(\tan^{-1}x)^2}{1+x^2} dx = \int_0^{\frac{\pi}{4}} u^2 du\) \(= \left[ \frac{u^3}{3} \right]_0^{\frac{\pi}{4}}\) \(= \frac{(\frac{\pi}{4})^3}{3} - \frac{0^3}{3}\) \(= \frac{\pi^3}{3 \cdot 4^3} = \frac{\pi^3}{3 \cdot 64} = \frac{\pi^3}{192}\) now, we have: \(\frac{16}{\pi^2} \int_0^1 \frac{(\tan^{-1}x)^2}{1+x^2} dx = \frac{16}{\pi^2} \cdot \frac{\pi^3}{192}\) \(= \frac{16 \pi^3}{192 \pi^2} = \frac{16 \pi}{192} = \frac{\pi}{12}\) so, \(\frac{16}{\pi^2} \int_0^1 \frac{(\tan^{-1}x)^2}{1+x^2} dx = \frac{\pi}{12}\) 🎉