int_(e^x) ((1+sinx)/(1+cosx)) এর মান হল- 

Explanation:

Another Explanation (5): সমাধান: আমরা জানি, \[ \frac{1 + \sin x}{1 + \cos x} = \frac{\sin^2 \frac{x}{2} + \cos^2 \frac{x}{2} + 2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos^2 \frac{x}{2}} = \frac{(\sin \frac{x}{2} + \cos \frac{x}{2})^2}{2 \cos^2 \frac{x}{2}} \] \[ = \frac{\sin^2 \frac{x}{2} + \cos^2 \frac{x}{2} + 2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos^2 \frac{x}{2}} = \frac{1 + 2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos^2 \frac{x}{2}} = \frac{1}{2 \cos^2 \frac{x}{2}} + \frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos^2 \frac{x}{2}} \] \[ = \frac{1}{2} \sec^2 \frac{x}{2} + \tan \frac{x}{2} \] এখন, \[ \int e^x \frac{1 + \sin x}{1 + \cos x} dx = \int e^x \left( \frac{1}{2} \sec^2 \frac{x}{2} + \tan \frac{x}{2} \right) dx \] আমরা জানি, \(\int e^x [f(x) + f'(x)] dx = e^x f(x) + c\) এখানে, \(f(x) = \tan \frac{x}{2}\) এবং \(f'(x) = \frac{1}{2} \sec^2 \frac{x}{2}\) সুতরাং, \[ \int e^x \left( \tan \frac{x}{2} + \frac{1}{2} \sec^2 \frac{x}{2} \right) dx = e^x \tan \frac{x}{2} + c \] অতএব, নির্ণেয় মান \(e^x \tan \frac{x}{2} + c\) 🥳🎉