I = ∫₀^π/4 (sin2θ/sin⁴θ + cos⁴θ)dθ এর মান কোনটি?

ধরি, \(I = \int_{0}^{\frac{\pi}{4}} \frac{\sin 2\theta}{\sin^4 \theta + \cos^4 \theta} d\theta\).

আমরা \(\sin^4 \theta + \cos^4 \theta\) কে লিখতে পারি:

\(\sin^4 \theta + \cos^4 \theta = (\sin^2 \theta + \cos^2 \theta)^2 - 2\sin^2 \theta \cos^2 \theta = 1 - 2\sin^2 \theta \cos^2 \theta = 1 - \frac{1}{2}(2\sin \theta \cos \theta)^2 = 1 - \frac{1}{2}\sin^2 2\theta\).

তাহলে,

\(I = \int_{0}^{\frac{\pi}{4}} \frac{\sin 2\theta}{1 - \frac{1}{2}\sin^2 2\theta} d\theta\).

ধরি, \(u = \sin 2\theta\). তাহলে, \(du = 2\cos 2\theta d\theta\). কিন্তু আমাদের শুধু \(\sin 2\theta\) আছে। তাই অন্য কিছু ধরতে হবে।

আবার ধরি, \(t = \sin^2 2\theta\). তাহলে, \(dt = 2\sin 2\theta \cdot 2\cos 2\theta d\theta = 4 \sin 2\theta \cos 2\theta d\theta = 2 \sin 4\theta d\theta \).

আমরা যদি \(u = \sin^2 2\theta\) ধরি, তাহলে \(du = 2 \sin 2\theta \cos 2\theta \cdot 2 d\theta = 2 \sin 4\theta d\theta\).

আবার প্রথমে ফিরে যাই:

\(I = \int_{0}^{\frac{\pi}{4}} \frac{\sin 2\theta}{1 - \frac{1}{2}\sin^2 2\theta} d\theta\).

ধরি, \(u = \sin^2 2\theta\). তাহলে, \(du = 2 \sin 2\theta \cos 2\theta (2 d\theta) = 2 \sin 4\theta d\theta \).

এখন, ধরি \(t = \sin 2\theta\). তাহলে, \(dt = 2 \cos 2\theta d\theta\). এটা কাজে লাগবে না।

আবার, \(I = \int_{0}^{\frac{\pi}{4}} \frac{\sin 2\theta}{\cos^4 \theta + \sin^4 \theta} d\theta = \int_{0}^{\frac{\pi}{4}} \frac{\sin 2\theta}{\cos^4 \theta (1 + \tan^4 \theta)} d\theta = \int_{0}^{\frac{\pi}{4}} \frac{2 \sin \theta \cos \theta}{\cos^4 \theta (1 + \tan^4 \theta)} d\theta\)

\(= \int_{0}^{\frac{\pi}{4}} \frac{2 \tan \theta}{\cos^2 \theta (1 + \tan^4 \theta)} d\theta = \int_{0}^{\frac{\pi}{4}} \frac{2 \tan \theta \sec^2 \theta}{1 + \tan^4 \theta} d\theta\)

ধরি, \(u = \tan^2 \theta\). তাহলে, \(du = 2 \tan \theta \sec^2 \theta d\theta\).

যখন \(\theta = 0\), \(u = \tan^2 0 = 0\). যখন \(\theta = \frac{\pi}{4}\), \(u = \tan^2 \frac{\pi}{4} = 1\).

সুতরাং, \(I = \int_{0}^{1} \frac{du}{1 + u^2} = [\arctan u]_{0}^{1} = \arctan 1 - \arctan 0 = \frac{\pi}{4} - 0 = \frac{\pi}{4}\).

অতএব, \(I = \frac{\pi}{4}\). 🎉🎉🎉