int_0^(pi/4)dx/(1+sinx)=?

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

সমাধান:

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

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

আমরা আরও জানি, \( 1+\tan^2(x/2) = \sec^2(x/2) \)। তাহলে,

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

যখন \( x = 0 \), তখন \( u = 1+\tan(0) = 1 \)।

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

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

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