int_0^(π/2) sqrt(1+sintheta)d theta   এর মান-

Explanation:

Another Explanation (5): সমাধান: আমরা জানি, \(sin \theta = (sin^2(\theta/2) + cos^2(\theta/2)) + 2sin(\theta/2)cos(\theta/2)\)। সুতরাং, \(1 + sin \theta = sin^2(\theta/2) + cos^2(\theta/2) + 2sin(\theta/2)cos(\theta/2) = (sin(\theta/2) + cos(\theta/2))^2\)। অতএব, \(\sqrt{1 + sin \theta} = \sqrt{(sin(\theta/2) + cos(\theta/2))^2} = |sin(\theta/2) + cos(\theta/2)|\)। যেহেতু \(0 \le \theta \le \pi/2\), তাই \(0 \le \theta/2 \le \pi/4\)। এই সীমার মধ্যে, \(sin(\theta/2) \ge 0\) এবং \(cos(\theta/2) \ge 0\)। সুতরাং, \(sin(\theta/2) + cos(\theta/2) \ge 0\)। তাহলে, \(\sqrt{1 + sin \theta} = sin(\theta/2) + cos(\theta/2)\)। এখন, আমরা ইন্টিগ্রালটি সমাধান করি: \(\int_0^{\pi/2} \sqrt{1 + sin \theta} \, d\theta = \int_0^{\pi/2} (sin(\theta/2) + cos(\theta/2)) \, d\theta\) = \(\int_0^{\pi/2} sin(\theta/2) \, d\theta + \int_0^{\pi/2} cos(\theta/2) \, d\theta\) = \([-2cos(\theta/2)]_0^{\pi/2} + [2sin(\theta/2)]_0^{\pi/2}\) = \((-2cos(\pi/4) - (-2cos(0))) + (2sin(\pi/4) - 2sin(0))\) = \((-2 \cdot \frac{1}{\sqrt{2}} + 2) + (2 \cdot \frac{1}{\sqrt{2}} - 0)\) = \(-\sqrt{2} + 2 + \sqrt{2}\) = \(2\) সুতরাং, \(\int_0^{\pi/2} \sqrt{1 + sin \theta} \, d\theta = 2\)। 🎉