int_0^(π/2)dx/(1+cosx)  এর মান নির্ণয় কর।

প্রশ্ন: \(\int_0^{\frac{\pi}{2}} \frac{dx}{1+\cos x}\) এর মান নির্ণয় কর।

সমাধান:

আমরা জানি, \(\cos x = \frac{1-\tan^2{\frac{x}{2}}}{1+\tan^2{\frac{x}{2}}}\)

সুতরাং, \( \int_0^{\frac{\pi}{2}} \frac{dx}{1+\cos x} = \int_0^{\frac{\pi}{2}} \frac{dx}{1+\frac{1-\tan^2{\frac{x}{2}}}{1+\tan^2{\frac{x}{2}}}}\)

\(= \int_0^{\frac{\pi}{2}} \frac{(1+\tan^2{\frac{x}{2}})dx}{1+\tan^2{\frac{x}{2}}+1-\tan^2{\frac{x}{2}}}\)

\(= \int_0^{\frac{\pi}{2}} \frac{\sec^2{\frac{x}{2}}dx}{2}\)

\(= \frac{1}{2} \int_0^{\frac{\pi}{2}} \sec^2{\frac{x}{2}}dx\)

ধরি, \(u = \frac{x}{2}\), সুতরাং \(du = \frac{1}{2}dx\), বা \(dx = 2du\)

যখন \(x = 0\), তখন \(u = 0\)

যখন \(x = \frac{\pi}{2}\), তখন \(u = \frac{\pi}{4}\)

অতএব, \(\frac{1}{2} \int_0^{\frac{\pi}{2}} \sec^2{\frac{x}{2}}dx = \frac{1}{2} \int_0^{\frac{\pi}{4}} \sec^2{u} (2du)\)

\(= \int_0^{\frac{\pi}{4}} \sec^2{u} du\)

\(= [\tan u]_0^{\frac{\pi}{4}}\)

\(= \tan{\frac{\pi}{4}} - \tan{0}\)

\(= 1 - 0\)

\(= 1\)

সুতরাং, \(\int_0^{\frac{\pi}{2}} \frac{dx}{1+\cos x} = 1\) 🎉