int_0^(pi/2)sqrt(1+sintheta)dt"theta=?

Explanation:

Another Explanation (5): ইন্টিগ্রেশনটির সমাধান নিচে দেওয়া হল: \[ \int_{0}^{\frac{\pi}{2}} \sqrt{1+\sin\theta} \, d\theta \] আমরা জানি, \(\sin\theta = \sin(2 \cdot \frac{\theta}{2}) = 2\sin\frac{\theta}{2}\cos\frac{\theta}{2}\) এবং \(1 = \sin^2\frac{\theta}{2} + \cos^2\frac{\theta}{2}\). সুতরাং, \[ 1+\sin\theta = \sin^2\frac{\theta}{2} + \cos^2\frac{\theta}{2} + 2\sin\frac{\theta}{2}\cos\frac{\theta}{2} = \left(\sin\frac{\theta}{2} + \cos\frac{\theta}{2}\right)^2 \] অতএব, \[ \sqrt{1+\sin\theta} = \sqrt{\left(\sin\frac{\theta}{2} + \cos\frac{\theta}{2}\right)^2} = \left|\sin\frac{\theta}{2} + \cos\frac{\theta}{2}\right| \] যেহেতু \(0 \le \theta \le \frac{\pi}{2}\), তাই \(0 \le \frac{\theta}{2} \le \frac{\pi}{4}\). এই সীমার মধ্যে \(\sin\frac{\theta}{2}\) এবং \(\cos\frac{\theta}{2}\) উভয়ই ধনাত্মক। সুতরাং, পরম মান চিহ্নটি সরানো যায়। \[ \sqrt{1+\sin\theta} = \sin\frac{\theta}{2} + \cos\frac{\theta}{2} \] এখন, ইন্টিগ্রেশনটি হবে: \[ \int_{0}^{\frac{\pi}{2}} \left(\sin\frac{\theta}{2} + \cos\frac{\theta}{2}\right) \, d\theta = \int_{0}^{\frac{\pi}{2}} \sin\frac{\theta}{2} \, d\theta + \int_{0}^{\frac{\pi}{2}} \cos\frac{\theta}{2} \, d\theta \] আমরা জানি, \(\int \sin(ax) \, dx = -\frac{1}{a}\cos(ax) + C\) এবং \(\int \cos(ax) \, dx = \frac{1}{a}\sin(ax) + C\). সুতরাং, \[ \int_{0}^{\frac{\pi}{2}} \sin\frac{\theta}{2} \, d\theta = \left[-2\cos\frac{\theta}{2}\right]_{0}^{\frac{\pi}{2}} = -2\cos\frac{\pi}{4} - (-2\cos 0) = -2\cdot\frac{1}{\sqrt{2}} + 2 = 2 - \sqrt{2} \] এবং \[ \int_{0}^{\frac{\pi}{2}} \cos\frac{\theta}{2} \, d\theta = \left[2\sin\frac{\theta}{2}\right]_{0}^{\frac{\pi}{2}} = 2\sin\frac{\pi}{4} - 2\sin 0 = 2\cdot\frac{1}{\sqrt{2}} - 0 = \sqrt{2} \] সুতরাং, \[ \int_{0}^{\frac{\pi}{2}} \left(\sin\frac{\theta}{2} + \cos\frac{\theta}{2}\right) \, d\theta = (2 - \sqrt{2}) + \sqrt{2} = 2 \] অতএব, \[ \int_{0}^{\frac{\pi}{2}} \sqrt{1+\sin\theta} \, d\theta = 2 \]