int_0^1(x)/(sqrt(1-x^2) )dx=? 

সমাধান:

ধরি, \(x = \sin(\theta)\). সুতরাং, \(dx = \cos(\theta) d\theta\).

যখন \(x = 0\), \(\theta = 0\). যখন \(x = 1\), \(\theta = \frac{\pi}{2}\).

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

\(= \int_0^{\frac{\pi}{2}} \frac{\sin(\theta)}{\sqrt{\cos^2(\theta)}} \cos(\theta) d\theta = \int_0^{\frac{\pi}{2}} \frac{\sin(\theta)}{\cos(\theta)} \cos(\theta) d\theta\)

\(= \int_0^{\frac{\pi}{2}} \sin(\theta) d\theta = [-\cos(\theta)]_0^{\frac{\pi}{2}}\)

\(= -\cos(\frac{\pi}{2}) - (-\cos(0)) = -0 + 1 = 1\). 🎉

সুতরাং, \(\int_0^1 \frac{x}{\sqrt{1-x^2}} dx = 1\).😊