int_(1/2)^1 (dx)/(xsqrt(4x^2-1))=? 

প্রশ্ন: \( \int_{1/2}^1 \frac{dx}{x\sqrt{4x^2-1}} = ? \)

সমাধান:

ধরি, \( 2x = \sec\theta \). তাহলে, \( 2 dx = \sec\theta \tan\theta d\theta \), অর্থাৎ \( dx = \frac{1}{2} \sec\theta \tan\theta d\theta \).

যখন \( x = \frac{1}{2} \), তখন \( \sec\theta = 2 \cdot \frac{1}{2} = 1 \), অর্থাৎ \( \theta = 0 \).

যখন \( x = 1 \), তখন \( \sec\theta = 2 \cdot 1 = 2 \), অর্থাৎ \( \theta = \frac{\pi}{3} \).

অতএব,

\( \int_{1/2}^1 \frac{dx}{x\sqrt{4x^2-1}} = \int_0^{\pi/3} \frac{\frac{1}{2}\sec\theta \tan\theta d\theta}{\frac{1}{2}\sec\theta \sqrt{\sec^2\theta - 1}} \)

\( = \int_0^{\pi/3} \frac{\sec\theta \tan\theta d\theta}{\sec\theta \sqrt{\tan^2\theta}} = \int_0^{\pi/3} \frac{\tan\theta d\theta}{\tan\theta} \)

\( = \int_0^{\pi/3} d\theta = [\theta]_0^{\pi/3} = \frac{\pi}{3} - 0 = \frac{\pi}{3} \). 🎉

সুতরাং, \( \int_{1/2}^1 \frac{dx}{x\sqrt{4x^2-1}} = \frac{\pi}{3} \). ✅