int_0^(pi/4) dx/(1+sinx)  এর মান কত? 

প্রশ্ন: \(\int_0^{\pi/4} \frac{dx}{1+\sin x}\) এর মান কত?

সমাধান:

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

সুতরাং, \(\int_0^{\pi/4} \frac{dx}{1+\sin x} = \int_0^{\pi/4} \frac{dx}{1+\frac{2\tan(x/2)}{1+\tan^2(x/2)}} = \int_0^{\pi/4} \frac{1+\tan^2(x/2)}{1+\tan^2(x/2)+2\tan(x/2)} dx\)

\(= \int_0^{\pi/4} \frac{\sec^2(x/2)}{(1+\tan(x/2))^2} dx\)

ধরি, \(u = 1+\tan(x/2)\). তাহলে, \(du = \frac{1}{2}\sec^2(x/2) dx\). সুতরাং, \(2du = \sec^2(x/2) dx\)

যখন \(x = 0\), \(u = 1+\tan(0) = 1\)

যখন \(x = \pi/4\), \(u = 1+\tan(\pi/8)\)

আমরা জানি, \(\tan(\pi/8) = \sqrt{2}-1\)

সুতরাং, \(u = 1+\sqrt{2}-1 = \sqrt{2}\)

তাহলে, \(\int_0^{\pi/4} \frac{\sec^2(x/2)}{(1+\tan(x/2))^2} dx = \int_1^{\sqrt{2}} \frac{2}{u^2} du = 2\int_1^{\sqrt{2}} u^{-2} du = 2\left[-\frac{1}{u}\right]_1^{\sqrt{2}}\)

\(= 2\left[-\frac{1}{\sqrt{2}} - (-1)\right] = 2\left[1-\frac{1}{\sqrt{2}}\right] = 2\left[1-\frac{\sqrt{2}}{2}\right] = 2-\sqrt{2}\)

অতএব, \(\int_0^{\pi/4} \frac{dx}{1+\sin x} = 2-\sqrt{2}\) 😊