int_0^(pi/4) 1/ (1+sin x) = ?

সমাধান:

আমরা \( \int_0^{\frac{\pi}{4}} \frac{1}{1 + \sin x} dx \) এর মান নির্ণয় করতে চাই।

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

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

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

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

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

যখন \( x = 0 \), \( u = \tan 0 = 0 \)।

যখন \( x = \frac{\pi}{4} \), \( u = \tan \frac{\pi}{8} \)। আমরা জানি, \( \tan \frac{\pi}{8} = \sqrt{2} - 1 \)।

সুতরাং, \( \int_0^{\frac{\pi}{4}} \frac{\sec^2 \frac{x}{2}}{(1 + \tan \frac{x}{2})^2} dx = \int_0^{\sqrt{2} - 1} \frac{2}{(1 + u)^2} du \)

\(= 2 \int_0^{\sqrt{2} - 1} (1 + u)^{-2} du \)

\(= 2 \left[ \frac{(1 + u)^{-1}}{-1} \right]_0^{\sqrt{2} - 1} \)

\(= -2 \left[ \frac{1}{1 + u} \right]_0^{\sqrt{2} - 1} \)

\(= -2 \left( \frac{1}{1 + \sqrt{2} - 1} - \frac{1}{1 + 0} \right) \)

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

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

\(= - \sqrt{2} + 2 \)

\(= 2 - \sqrt{2} \)

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