int_(π/2)^π (1+sin2θ)dθ   এর মান কত?

দেয়া আছে, \( \int_{\frac{\pi}{2}}^{\pi} (1+\sin 2\theta) d\theta \) এর মান নির্ণয় করতে হবে। 🧐

আমরা জানি, \( \int (f(x) + g(x)) dx = \int f(x) dx + \int g(x) dx \) এবং \( \int \sin(ax) dx = -\frac{1}{a} \cos(ax) + C \)। 🤓

সুতরাং,

\( \int_{\frac{\pi}{2}}^{\pi} (1+\sin 2\theta) d\theta = \int_{\frac{\pi}{2}}^{\pi} 1 d\theta + \int_{\frac{\pi}{2}}^{\pi} \sin 2\theta d\theta \) 😊

এখন, \( \int_{\frac{\pi}{2}}^{\pi} 1 d\theta = [\theta]_{\frac{\pi}{2}}^{\pi} = \pi - \frac{\pi}{2} = \frac{\pi}{2} \) 🥳

এবং, \( \int_{\frac{\pi}{2}}^{\pi} \sin 2\theta d\theta = [-\frac{1}{2} \cos 2\theta]_{\frac{\pi}{2}}^{\pi} = -\frac{1}{2} (\cos 2\pi - \cos \pi) = -\frac{1}{2} (1 - (-1)) = -\frac{1}{2} (2) = -1 \) 🤩

তাহলে, \( \int_{\frac{\pi}{2}}^{\pi} (1+\sin 2\theta) d\theta = \frac{\pi}{2} - 1 \) 🎉

অতএব, \( \int_{\frac{\pi}{2}}^{\pi} (1+\sin 2\theta) d\theta = \frac{\pi}{2} - 1 \)। 💯