int_0^3dx/(sqrt(3x-x^2)=? 

Explanation:

Another Explanation (5): সমাধান: \[ \int_0^3 \frac{dx}{\sqrt{3x-x^2}} \] প্রথমে, ইন্টিগ্রান্ডটিকে সরল করা যাক: \[ 3x - x^2 = -(x^2 - 3x) = -\left(x^2 - 3x + \left(\frac{3}{2}\right)^2 - \left(\frac{3}{2}\right)^2\right) = -\left(\left(x - \frac{3}{2}\right)^2 - \frac{9}{4}\right) = \frac{9}{4} - \left(x - \frac{3}{2}\right)^2 \] সুতরাং, ইন্টিগ্রালটি হবে: \[ \int_0^3 \frac{dx}{\sqrt{\frac{9}{4} - \left(x - \frac{3}{2}\right)^2}} \] এখন, ধরা যাক \(x - \frac{3}{2} = \frac{3}{2}\sin\theta\). তাহলে, \(dx = \frac{3}{2}\cos\theta d\theta\). যখন \(x = 0\), \(\frac{3}{2}\sin\theta = -\frac{3}{2}\), সুতরাং \(\sin\theta = -1\), অর্থাৎ \(\theta = -\frac{\pi}{2}\). যখন \(x = 3\), \(\frac{3}{2}\sin\theta = \frac{3}{2}\), সুতরাং \(\sin\theta = 1\), অর্থাৎ \(\theta = \frac{\pi}{2}\). সুতরাং, ইন্টিগ্রালটি হবে: \[ \int_{-\pi/2}^{\pi/2} \frac{\frac{3}{2}\cos\theta d\theta}{\sqrt{\frac{9}{4} - \frac{9}{4}\sin^2\theta}} = \int_{-\pi/2}^{\pi/2} \frac{\frac{3}{2}\cos\theta d\theta}{\frac{3}{2}\sqrt{1 - \sin^2\theta}} = \int_{-\pi/2}^{\pi/2} \frac{\frac{3}{2}\cos\theta d\theta}{\frac{3}{2}\cos\theta} = \int_{-\pi/2}^{\pi/2} d\theta \] \[ = [\theta]_{-\pi/2}^{\pi/2} = \frac{\pi}{2} - \left(-\frac{\pi}{2}\right) = \frac{\pi}{2} + \frac{\pi}{2} = \pi \] সুতরাং, \[ \int_0^3 \frac{dx}{\sqrt{3x-x^2}} = \pi \] 🎉🥳