intdx/(sqrt(2ax-x^2))=? 

Explanation:

Another Explanation (5): সমাধান: \[ \int \frac{dx}{\sqrt{2ax - x^2}} = ? \] আমরা প্রথমে \(2ax - x^2\) কে পূর্ণ বর্গ আকারে লিখি: \[ 2ax - x^2 = -(x^2 - 2ax) = -(x^2 - 2ax + a^2 - a^2) = a^2 - (x - a)^2 \] তাহলে, \[ \int \frac{dx}{\sqrt{2ax - x^2}} = \int \frac{dx}{\sqrt{a^2 - (x - a)^2}} \] এখন, \(x - a = a\sin\theta\) ধরি। তাহলে, \(dx = a\cos\theta d\theta\). সুতরাং, \[ \int \frac{dx}{\sqrt{a^2 - (x - a)^2}} = \int \frac{a\cos\theta d\theta}{\sqrt{a^2 - a^2\sin^2\theta}} = \int \frac{a\cos\theta d\theta}{\sqrt{a^2(1 - \sin^2\theta)}} \] \[ = \int \frac{a\cos\theta d\theta}{a\cos\theta} = \int d\theta = \theta + C \] যেহেতু \(x - a = a\sin\theta\), তাই \(\sin\theta = \frac{x - a}{a}\), সুতরাং \(\theta = \sin^{-1}\left(\frac{x - a}{a}\right)\). অতএব, \[ \int \frac{dx}{\sqrt{2ax - x^2}} = \sin^{-1}\left(\frac{x - a}{a}\right) + C \] সুতরাং নির্ণেয় সমাধান: 🎉🎉