int_0^(pi/4)(sin2theta)/(sin^4theta+cos^4theta)d theta

এর মান কোনটি?

ধরি, \( I = \int_0^{\pi/4} \frac{\sin 2\theta}{\sin^4 \theta + \cos^4 \theta} d\theta \)

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

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

সুতরাং,

\( I = \int_0^{\pi/4} \frac{\sin 2\theta}{1 - \frac{1}{2}\sin^2 2\theta} d\theta \)

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

ধরি, \( t = \sin^2 2\theta \), তাহলে \( dt = 2\sin 2\theta \cdot 2\cos 2\theta d\theta = 2\sin 4\theta d\theta \). এটা কাজে লাগবে না।

আবার ধরি, \( u = \sin^2 2\theta \). যখন \( \theta = 0 \), \( u = \sin^2 0 = 0 \). যখন \( \theta = \pi/4 \), \( u = \sin^2 (\pi/2) = 1 \).

\( I = \int_0^{\pi/4} \frac{\sin 2\theta}{1 - \frac{1}{2}\sin^2 2\theta} d\theta = 2 \int_0^{\pi/4} \frac{\frac{1}{2}\sin 2\theta}{1 - \frac{1}{2}\sin^2 2\theta} d\theta \)

এখন, ধরি \( u = 1 - \frac{1}{2}\sin^2 2\theta \). তাহলে, \( du = -\frac{1}{2} \cdot 2 \sin 2\theta \cdot 2 \cos 2\theta d\theta = -\sin 2\theta (2\cos 2\theta) d\theta \). 🤔 এটা কাজ করছে না।

আবার, \( I = \int_0^{\pi/4} \frac{\sin 2\theta}{\cos^4 \theta + \sin^4 \theta} d\theta = \int_0^{\pi/4} \frac{\sin 2\theta}{\cos^4 \theta (1 + \tan^4 \theta)} d\theta = \int_0^{\pi/4} \frac{\frac{\sin 2\theta}{\cos^2 \theta}}{\cos^2 \theta (1 + \tan^4 \theta)} d\theta = \int_0^{\pi/4} \frac{2\tan \theta \sec^2 \theta}{1 + \tan^4 \theta} d\theta \)

ধরি, \( u = \tan^2 \theta \), তাহলে \( du = 2\tan \theta \sec^2 \theta d\theta \). যখন \( \theta = 0 \), \( u = 0 \). যখন \( \theta = \pi/4 \), \( u = 1 \).

সুতরাং, \( I = \int_0^1 \frac{du}{1 + u^2} = [\arctan u]_0^1 = \arctan 1 - \arctan 0 = \frac{\pi}{4} - 0 = \frac{\pi}{4} \) 🎉