int_0^(1/2)1/sqrt(1-3x^2)dx=P হলে , P এর মান কত?

Type explanation here...

ধরি, \(I = \int_{0}^{\frac{1}{2}} \frac{1}{\sqrt{1-3x^2}} dx\)

এখন, \(3x^2 = \sin^2{\theta}\) ধরি।

তাহলে, \(\sqrt{3}x = \sin{\theta}\)

অতএব, \(\sqrt{3} dx = \cos{\theta} d\theta\)

সুতরাং, \(dx = \frac{\cos{\theta}}{\sqrt{3}} d\theta\)

যখন \(x = 0\), \(\sin{\theta} = 0\), সুতরাং \(\theta = 0\)

যখন \(x = \frac{1}{2}\), \(\sin{\theta} = \sqrt{3} \cdot \frac{1}{2} = \frac{\sqrt{3}}{2}\), সুতরাং \(\theta = \frac{\pi}{3}\)

তাহলে, \(I = \int_{0}^{\frac{\pi}{3}} \frac{1}{\sqrt{1-\sin^2{\theta}}} \cdot \frac{\cos{\theta}}{\sqrt{3}} d\theta\)

\(= \int_{0}^{\frac{\pi}{3}} \frac{1}{\sqrt{\cos^2{\theta}}} \cdot \frac{\cos{\theta}}{\sqrt{3}} d\theta\)

\(= \int_{0}^{\frac{\pi}{3}} \frac{1}{\cos{\theta}} \cdot \frac{\cos{\theta}}{\sqrt{3}} d\theta\)

\(= \int_{0}^{\frac{\pi}{3}} \frac{1}{\sqrt{3}} d\theta\)

\(= \frac{1}{\sqrt{3}} \int_{0}^{\frac{\pi}{3}} d\theta\)

\(= \frac{1}{\sqrt{3}} [\theta]_{0}^{\frac{\pi}{3}}\)

\(= \frac{1}{\sqrt{3}} \left( \frac{\pi}{3} - 0 \right)\)

\(= \frac{\pi}{3\sqrt{3}}\)

অতএব, \(P = \frac{\pi}{3\sqrt{3}}\) 🎉