int_0^(pi/2)(sinx)/(sinx+cosx)dx=? 

Explanation:

Another Explanation (5): সমাধান: ধরি, \(I = \int_{0}^{\frac{\pi}{2}} \frac{\sin x}{\sin x + \cos x} dx\) 😊 এখন, \(x = \frac{\pi}{2} - t\) প্রতিস্থাপন করি। সুতরাং, \(dx = -dt\) হবে। যখন \(x = 0\), তখন \(t = \frac{\pi}{2}\) এবং যখন \(x = \frac{\pi}{2}\), তখন \(t = 0\)। তাহলে, \(I = \int_{\frac{\pi}{2}}^{0} \frac{\sin(\frac{\pi}{2} - t)}{\sin(\frac{\pi}{2} - t) + \cos(\frac{\pi}{2} - t)} (-dt)\) \(I = \int_{0}^{\frac{\pi}{2}} \frac{\cos t}{\cos t + \sin t} dt\) \(I = \int_{0}^{\frac{\pi}{2}} \frac{\cos x}{\cos x + \sin x} dx\) ✨ সুতরাং, \(I = \int_{0}^{\frac{\pi}{2}} \frac{\cos x}{\sin x + \cos x} dx\) এখন, আমরা প্রথম সমীকরণ এবং এই নতুন সমীকরণ যোগ করি: \(I + I = \int_{0}^{\frac{\pi}{2}} \frac{\sin x}{\sin x + \cos x} dx + \int_{0}^{\frac{\pi}{2}} \frac{\cos x}{\sin x + \cos x} dx\) \(2I = \int_{0}^{\frac{\pi}{2}} \frac{\sin x + \cos x}{\sin x + \cos x} dx\) \(2I = \int_{0}^{\frac{\pi}{2}} 1 dx\) 🎉 \(2I = [x]_{0}^{\frac{\pi}{2}}\) \(2I = \frac{\pi}{2} - 0\) \(2I = \frac{\pi}{2}\) সুতরাং, \(I = \frac{\pi}{4}\) 🎈 অতএব, \(\int_{0}^{\frac{\pi}{2}} \frac{\sin x}{\sin x + \cos x} dx = \frac{\pi}{4}\) 😎