int 1/(1+cosx)dx=? 

Explanation:

Another Explanation (5): সমাধান: আমরা জানি, \( \cos x = \frac{1 - \tan^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2}} \) সুতরাং, \( \int \frac{1}{1 + \cos x} dx = \int \frac{1}{1 + \frac{1 - \tan^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2}}} dx \) \( = \int \frac{1}{\frac{1 + \tan^2 \frac{x}{2} + 1 - \tan^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2}}} dx \) \( = \int \frac{1 + \tan^2 \frac{x}{2}}{2} dx \) \( = \frac{1}{2} \int (1 + \tan^2 \frac{x}{2}) dx \) আমরা জানি, \( 1 + \tan^2 \theta = \sec^2 \theta \) \( = \frac{1}{2} \int \sec^2 \frac{x}{2} dx \) এখন, \( \int \sec^2 ax \, dx = \frac{\tan ax}{a} + C \) এই সূত্র ব্যবহার করে পাই, \( = \frac{1}{2} \cdot \frac{\tan \frac{x}{2}}{\frac{1}{2}} + C \) \( = \tan \frac{x}{2} + C \) অতএব, \( \int \frac{1}{1 + \cos x} dx = \tan \frac{x}{2} + C \) 🎉