int_0^1sqrt(1-x^2)dx=? 

সমাধান:

ধরি, \(x = \sin\theta\)

তাহলে, \(dx = \cos\theta d\theta\)

যখন \(x = 0\), \(\theta = 0\)

যখন \(x = 1\), \(\theta = \pi/2\)

সুতরাং, \(\int_0^1 \sqrt{1-x^2} dx = \int_0^{\pi/2} \sqrt{1-\sin^2\theta} \cos\theta d\theta\)

\(= \int_0^{\pi/2} \cos^2\theta d\theta\)

\(= \int_0^{\pi/2} \frac{1+\cos2\theta}{2} d\theta\)

\(= \frac{1}{2} \int_0^{\pi/2} (1+\cos2\theta) d\theta\)

\(= \frac{1}{2} \left[ \theta + \frac{\sin2\theta}{2} \right]_0^{\pi/2}\)

\(= \frac{1}{2} \left[ \frac{\pi}{2} + \frac{\sin\pi}{2} - (0 + \frac{\sin0}{2}) \right]\)

\(= \frac{1}{2} \left[ \frac{\pi}{2} + 0 - 0 \right]\)

\(= \frac{\pi}{4}\) 🎉

অতএব, \(\int_0^1 \sqrt{1-x^2} dx = \frac{\pi}{4}\) 🥳