int_0^(pi/4)dx/(1+sinx)=? 

প্রশ্ন: \( \int_{0}^{\pi/4} \frac{dx}{1+\sin x} = ? \)

সমাধান:

আমরা জানি, \( \sin x = \frac{2\tan(x/2)}{1+\tan^2(x/2)} \). ধরি, \( t = \tan(x/2) \). তাহলে, \( dt = \frac{1}{2}\sec^2(x/2) dx = \frac{1}{2}(1+\tan^2(x/2))dx = \frac{1}{2}(1+t^2)dx \). সুতরাং, \( dx = \frac{2dt}{1+t^2} \).

যখন \( x = 0 \), তখন \( t = \tan(0/2) = 0 \). যখন \( x = \pi/4 \), তখন \( t = \tan(\pi/8) \). আমরা জানি, \( \tan(\pi/8) = \sqrt{2}-1 \). সুতরাং, নতুন সীমা \( 0 \) থেকে \( \sqrt{2}-1 \).

তাহলে, \( \int_{0}^{\pi/4} \frac{dx}{1+\sin x} = \int_{0}^{\sqrt{2}-1} \frac{1}{1+\frac{2t}{1+t^2}} \cdot \frac{2dt}{1+t^2} = \int_{0}^{\sqrt{2}-1} \frac{2dt}{1+t^2+2t} = \int_{0}^{\sqrt{2}-1} \frac{2dt}{(1+t)^2} \).

এখন, \( \int \frac{2}{(1+t)^2} dt = 2 \int (1+t)^{-2} dt = 2 \cdot \frac{(1+t)^{-1}}{-1} + C = -\frac{2}{1+t} + C \).

সুতরাং, \( \int_{0}^{\sqrt{2}-1} \frac{2dt}{(1+t)^2} = \left[ -\frac{2}{1+t} \right]_{0}^{\sqrt{2}-1} = -\frac{2}{1+\sqrt{2}-1} - \left( -\frac{2}{1+0} \right) = -\frac{2}{\sqrt{2}} + 2 = -\sqrt{2} + 2 = 2-\sqrt{2} \).

অতএব, \( \int_{0}^{\pi/4} \frac{dx}{1+\sin x} = 2-\sqrt{2} \). 🎉