int sin(1/x)/x^2dx =?

Explanation:

Another Explanation (5): সমাধান: ধরি, \(u = \frac{1}{x}\) তাহলে, \(\frac{du}{dx} = -\frac{1}{x^2}\) সুতরাং, \(du = -\frac{1}{x^2} dx\) অতএব, \(\frac{1}{x^2} dx = -du\) এখন, \(\int \frac{\sin(\frac{1}{x})}{x^2} dx = \int \sin(u) (-du) = -\int \sin(u) du\) আমরা জানি, \(\int \sin(u) du = -\cos(u) + c\) , যেখানে c একটি সমাকলন ধ্রুবক। সুতরাং, \(-\int \sin(u) du = -(-\cos(u)) + c = \cos(u) + c\) u এর মান বসিয়ে পাই, \(\cos(\frac{1}{x}) + c\) অতএব, \(\int \frac{\sin(\frac{1}{x})}{x^2} dx = \cos(\frac{1}{x}) + c\) 🎉🎉🎉