Evaluated: int_(-3/4)^(3/4)( dx) / (√(9-4x^2) 

Explanation:

Another Explanation (5): সমাধান: \[ \int_{-\frac{3}{4}}^{\frac{3}{4}} \frac{dx}{\sqrt{9-4x^2}} \] ধরি, \( x = \frac{3}{2} \sin(\theta) \). তাহলে, \( dx = \frac{3}{2} \cos(\theta) d\theta \). 😃 যখন \( x = -\frac{3}{4} \), \[ -\frac{3}{4} = \frac{3}{2} \sin(\theta) \implies \sin(\theta) = -\frac{1}{2} \implies \theta = -\frac{\pi}{6} \] যখন \( x = \frac{3}{4} \), \[ \frac{3}{4} = \frac{3}{2} \sin(\theta) \implies \sin(\theta) = \frac{1}{2} \implies \theta = \frac{\pi}{6} \] সুতরাং, ইন্টিগ্রালটি হবে: \[ \int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} \frac{\frac{3}{2} \cos(\theta) d\theta}{\sqrt{9 - 4(\frac{9}{4} \sin^2(\theta))}} = \int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} \frac{\frac{3}{2} \cos(\theta) d\theta}{\sqrt{9 - 9\sin^2(\theta)}} \] \[ = \int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} \frac{\frac{3}{2} \cos(\theta) d\theta}{3\sqrt{1 - \sin^2(\theta)}} = \int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} \frac{\frac{3}{2} \cos(\theta) d\theta}{3\cos(\theta)} \] \[ = \int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} \frac{1}{2} d\theta = \frac{1}{2} [\theta]_{-\frac{\pi}{6}}^{\frac{\pi}{6}} = \frac{1}{2} \left( \frac{\pi}{6} - \left(-\frac{\pi}{6}\right) \right) \] \[ = \frac{1}{2} \left( \frac{\pi}{6} + \frac{\pi}{6} \right) = \frac{1}{2} \left( \frac{2\pi}{6} \right) = \frac{\pi}{6} \] অতএব, \[ \int_{-\frac{3}{4}}^{\frac{3}{4}} \frac{dx}{\sqrt{9-4x^2}} = \frac{\pi}{6} \] 🎉🎉