I=int_0^(π/4) (Sin2theta)/(Sin^4theta+Cos^4theta) d theta=? 

Explanation:

Another Explanation (5): আমরা \(I\) এর মান নির্ণয় করব: \[ I = \int_{0}^{\frac{\pi}{4}} \frac{\sin 2\theta}{\sin^4 \theta + \cos^4 \theta} d\theta \] প্রথমে হরটিকে সরল করা যাক: \[ \begin{aligned} \sin^4 \theta + \cos^4 \theta &= (\sin^2 \theta + \cos^2 \theta)^2 - 2\sin^2 \theta \cos^2 \theta \\ &= 1 - 2\sin^2 \theta \cos^2 \theta \\ &= 1 - \frac{1}{2} (2 \sin \theta \cos \theta)^2 \\ &= 1 - \frac{1}{2} \sin^2 2\theta \end{aligned} \] তাহলে, \[ I = \int_{0}^{\frac{\pi}{4}} \frac{\sin 2\theta}{1 - \frac{1}{2} \sin^2 2\theta} d\theta \] এখন, ধরি \(u = \cos 2\theta\), তাহলে \(-du = 2\sin 2\theta d\theta\), অর্থাৎ \(\sin 2\theta d\theta = -\frac{1}{2} du\). সীমা পরিবর্তন করি: যখন \(\theta = 0\), \(u = \cos 0 = 1\). যখন \(\theta = \frac{\pi}{4}\), \(u = \cos \frac{\pi}{2} = 0\). সুতরাং, \[ \begin{aligned} I &= \int_{1}^{0} \frac{-\frac{1}{2} du}{1 - \frac{1}{2}(1 - u^2)} \\ &= -\frac{1}{2} \int_{1}^{0} \frac{du}{1 - \frac{1}{2} + \frac{1}{2}u^2} \\ &= -\frac{1}{2} \int_{1}^{0} \frac{du}{\frac{1}{2} + \frac{1}{2}u^2} \\ &= -\int_{1}^{0} \frac{du}{1 + u^2} \\ &= -[\tan^{-1} u]_{1}^{0} \\ &= -(\tan^{-1} 0 - \tan^{-1} 1) \\ &= -(0 - \frac{\pi}{4}) \\ &= \frac{\pi}{4} \end{aligned} \] অতএব, \[ I = \frac{\pi}{4} \] ✅