int_0^((pi^2)/4)sinsqrtxdx =কত?

Explanation:

Another Explanation (5): সমাধান: ধরি, \(u = \sqrt{x}\), সুতরাং \(x = u^2\) এবং \(dx = 2u \, du\). সীমা পরিবর্তন করে পাই, যখন \(x = 0\), \(u = \sqrt{0} = 0\) এবং যখন \(x = \frac{\pi^2}{4}\), \(u = \sqrt{\frac{\pi^2}{4}} = \frac{\pi}{2}\). সুতরাং, \(\int_0^{\frac{\pi^2}{4}} \sin(\sqrt{x}) \, dx = \int_0^{\frac{\pi}{2}} \sin(u) \cdot 2u \, du = 2 \int_0^{\frac{\pi}{2}} u \sin(u) \, du\) এখন, আমরা ইন্টিগ্রেশন বাই পার্টস ব্যবহার করি: \(\int v \, dw = vw - \int w \, dv\) ধরি, \(v = u\) এবং \(dw = \sin(u) \, du\). তাহলে, \(dv = du\) এবং \(w = -\cos(u)\). সুতরাং, \(2 \int_0^{\frac{\pi}{2}} u \sin(u) \, du = 2 \left[ -u \cos(u) \Big|_0^{\frac{\pi}{2}} - \int_0^{\frac{\pi}{2}} (-\cos(u)) \, du \right]\) \(= 2 \left[ -u \cos(u) \Big|_0^{\frac{\pi}{2}} + \int_0^{\frac{\pi}{2}} \cos(u) \, du \right]\) \(= 2 \left[ -\frac{\pi}{2} \cos\left(\frac{\pi}{2}\right) - (-0 \cdot \cos(0)) + \sin(u) \Big|_0^{\frac{\pi}{2}} \right]\) \(= 2 \left[ -\frac{\pi}{2} \cdot 0 + 0 + \sin\left(\frac{\pi}{2}\right) - \sin(0) \right]\) \(= 2 \left[ 0 + 0 + 1 - 0 \right]\) \(= 2 \cdot 1 = 2\) সুতরাং, \(\int_0^{\frac{\pi^2}{4}} \sin(\sqrt{x}) \, dx = 2\) 🎉