int_0^(2a) dx/(sqrt(2ax-x^2) ) =? 

প্রশ্ন: \( \int_0^{2a} \frac{dx}{\sqrt{2ax-x^2}} = ? \)

সমাধান:

ধরি, \(I = \int_0^{2a} \frac{dx}{\sqrt{2ax-x^2}}\)

আমরা লিখতে পারি,

\(I = \int_0^{2a} \frac{dx}{\sqrt{a^2 - (x^2 - 2ax + a^2)}}\)

\(I = \int_0^{2a} \frac{dx}{\sqrt{a^2 - (x-a)^2}}\)

এখন, ধরি \(x-a = a\sin\theta\), সুতরাং \(dx = a\cos\theta d\theta\)

যখন \(x = 0\), তখন \(a\sin\theta = -a \implies \sin\theta = -1 \implies \theta = -\frac{\pi}{2}\)

যখন \(x = 2a\), তখন \(a\sin\theta = a \implies \sin\theta = 1 \implies \theta = \frac{\pi}{2}\)

সুতরাং,

\(I = \int_{-\pi/2}^{\pi/2} \frac{a\cos\theta d\theta}{\sqrt{a^2 - a^2\sin^2\theta}}\)

\(I = \int_{-\pi/2}^{\pi/2} \frac{a\cos\theta d\theta}{a\sqrt{1 - \sin^2\theta}}\)

\(I = \int_{-\pi/2}^{\pi/2} \frac{a\cos\theta d\theta}{a\cos\theta}\)

\(I = \int_{-\pi/2}^{\pi/2} d\theta\)

\(I = [\theta]_{-\pi/2}^{\pi/2}\)

\(I = \frac{\pi}{2} - (-\frac{\pi}{2})\)

\(I = \frac{\pi}{2} + \frac{\pi}{2}\)

\(I = \pi\)

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