int_-2^2 dx/(x^2+4) =  কত?

সমাধান:

আমরা জানি, \( \int \frac{1}{x^2 + a^2} dx = \frac{1}{a} \tan^{-1}(\frac{x}{a}) + C \)

এখানে, \( \int_{-2}^{2} \frac{1}{x^2 + 4} dx = \int_{-2}^{2} \frac{1}{x^2 + 2^2} dx \)

সুতরাং, \( \int_{-2}^{2} \frac{1}{x^2 + 4} dx = \frac{1}{2} \left[ \tan^{-1}(\frac{x}{2}) \right]_{-2}^{2} \)

এখন, মান বসিয়ে পাই,

\( = \frac{1}{2} \left[ \tan^{-1}(\frac{2}{2}) - \tan^{-1}(\frac{-2}{2}) \right] \)

\( = \frac{1}{2} \left[ \tan^{-1}(1) - \tan^{-1}(-1) \right] \)

আমরা জানি, \( \tan^{-1}(1) = \frac{\pi}{4} \) এবং \( \tan^{-1}(-1) = -\frac{\pi}{4} \)

অতএব,

\( = \frac{1}{2} \left[ \frac{\pi}{4} - (-\frac{\pi}{4}) \right] \)

\( = \frac{1}{2} \left[ \frac{\pi}{4} + \frac{\pi}{4} \right] \)

\( = \frac{1}{2} \left[ \frac{2\pi}{4} \right] \)

\( = \frac{\pi}{4} \)

সুতরাং, \( \int_{-2}^{2} \frac{dx}{x^2 + 4} = \frac{\pi}{4} \) 🥳