int_0^(pi/2) cosx/(9-sin^2x) dx এর মান কোনটি ? 

ধরি, \(I = \int_0^{\pi/2} \frac{\cos x}{9 - \sin^2 x} dx\)

এখানে, \(\sin x = t\) ধরলে, \(\cos x dx = dt\) হবে।

সুতরাং, যখন \(x = 0\), তখন \(t = \sin 0 = 0\) এবং যখন \(x = \pi/2\), তখন \(t = \sin(\pi/2) = 1\).

তাহলে, \(I = \int_0^1 \frac{dt}{9 - t^2}\)

আমরা জানি, \(\int \frac{dx}{a^2 - x^2} = \frac{1}{2a} \ln \left| \frac{a+x}{a-x} \right| + C\)

অতএব, \(I = \int_0^1 \frac{dt}{3^2 - t^2} = \frac{1}{2 \cdot 3} \left[ \ln \left| \frac{3+t}{3-t} \right| \right]_0^1\)

\(I = \frac{1}{6} \left[ \ln \left| \frac{3+1}{3-1} \right| - \ln \left| \frac{3+0}{3-0} \right| \right]\)

\(I = \frac{1}{6} \left[ \ln \frac{4}{2} - \ln 1 \right]\)

\(I = \frac{1}{6} \left[ \ln 2 - 0 \right]\)

\(I = \frac{1}{6} \ln 2\)

সুতরাং, \(\int_0^{\pi/2} \frac{\cos x}{9 - \sin^2 x} dx = \frac{1}{6} \ln 2\). 🎉