int_0^(lamda/4)(sin2xdx)/(sin^4x+cos^4x)=? 

প্রশ্ন: \( \int_{0}^{\lambda/4} \frac{\sin 2x}{\sin^4 x + \cos^4 x} dx = ? \)

উত্তর: \(\lambda/4\)

সমাধান:

ধরি, \( I = \int_{0}^{\lambda/4} \frac{\sin 2x}{\sin^4 x + \cos^4 x} dx \)

আমরা জানি, \(\sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - 2\sin^2 x \cos^2 x = 1 - \frac{1}{2}(2\sin x \cos x)^2 = 1 - \frac{1}{2}\sin^2 2x\)

সুতরাং, \( I = \int_{0}^{\lambda/4} \frac{\sin 2x}{1 - \frac{1}{2}\sin^2 2x} dx \)

এখন, ধরি \( u = \sin^2 2x \). তাহলে, \( du = 2\sin 2x \cdot \cos 2x \cdot 2 dx = 4 \sin 2x \cos 2x dx \) অথবা, \(\sin 4x dx \)

পরিবর্তন করে পাই, \( du = 4 \sin 2x \cos 2x dx \) অথবা, \( \frac{1}{2}du = 2 \sin 2x \cos 2x dx = \sin 4x \)

আবার ধরি, \( t = \sin^2 2x \), সুতরাং \(dt = 2\sin(2x)\cos(2x)*2 dx = 2\sin(4x) dx \) অথবা, \( \sin2x dx = \frac{dt}{2\cos(2x)} \). এই প্রতিস্থাপনটি সরাসরি ব্যবহার করা কঠিন।

অন্যভাবে করি, ধরি \( t = \sin^2(2x) \) তাহলে, \(dt = 2\sin(2x)\cos(2x)2 dx = 4\sin(2x)\cos(2x) dx = 2\sin(4x)dx \) সুতরাং \( \sin 2x dx = \frac{dt}{4\cos(2x)} \)

আচ্ছা, \( \sin^4 x + \cos^4 x = (sin^2 x + cos^2 x)^2 - 2sin^2 x cos^2 x = 1 - \frac{1}{2}sin^2(2x) \)

ধরি, \(u = cos2x \) , তাহলে \(du = -2sin2x dx \) সুতরাং \(sin2x dx = -\frac{1}{2}du \) যদি \( x = 0\) হয়, \(u = cos(0) = 1 \) যদি \( x = \frac{\lambda}{4} \) হয়, \(u = cos(\frac{\lambda}{2}) \) \( I = \int_{1}^{cos(\lambda/2)} \frac{-1/2}{sin^4 x + cos^4 x}du \)

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

সুতরাং, \( I = \int_{0}^{\lambda/4} \frac{\sin 2x}{\frac{1}{2}(1 + \cos^2 2x)} dx = 2\int_{0}^{\lambda/4} \frac{\sin 2x}{1 + \cos^2 2x} dx \)

ধরি, \(v = \cos 2x\) , \(dv = -2\sin 2x dx \), সুতরাং \( sin 2x dx = -\frac{dv}{2} \) যখন \(x = 0 \), \(v = 1 \), যখন \(x = \lambda/4 \), \(v = \cos (\lambda/2) \) \(I = 2\int_{1}^{cos(\lambda/2)} \frac{-1/2}{1+v^2}dv = -\int_{1}^{cos(\lambda/2)} \frac{1}{1+v^2}dv = \int_{cos(\lambda/2)}^{1} \frac{1}{1+v^2}dv = [tan^{-1}(v)]_{cos(\lambda/2)}^{1} = tan^{-1}(1) - tan^{-1}(cos(\lambda/2)) = \frac{\pi}{4} - tan^{-1}(cos(\lambda/2)) \)

যদি \( \lambda = \pi \), \( I = \frac{\pi}{4} - tan^{-1}(cos(\pi/2)) = \frac{\pi}{4} - tan^{-1}(0) = \frac{\pi}{4} \). তবে উত্তর \(\lambda/4\) হতে হলে \(\lambda = \pi\) হতে হবে। 🤔