int_0^(π/4) (costheta)/(cos^2theta) d theta =? 

প্রশ্নটি হলো: \(\int_{0}^{\frac{\pi}{4}} \frac{\cos(\theta)}{\cos^2(\theta)} d\theta = ?\)

সমাধান: \(\int_{0}^{\frac{\pi}{4}} \frac{\cos(\theta)}{\cos^2(\theta)} d\theta\) = \(\int_{0}^{\frac{\pi}{4}} \frac{1}{\cos(\theta)} d\theta\) = \(\int_{0}^{\frac{\pi}{4}} \sec(\theta) d\theta\)

আমরা জানি, \(\int \sec(x) dx = \ln|\sec(x) + \tan(x)| + C\)

সুতরাং, \(\int_{0}^{\frac{\pi}{4}} \sec(\theta) d\theta = \left[ \ln|\sec(\theta) + \tan(\theta)| \right]_{0}^{\frac{\pi}{4}}\)

এখন, \(\theta = \frac{\pi}{4}\) হলে, \(\sec(\frac{\pi}{4}) = \sqrt{2}\) এবং \(\tan(\frac{\pi}{4}) = 1\) সুতরাং, \(\ln|\sec(\frac{\pi}{4}) + \tan(\frac{\pi}{4})| = \ln|\sqrt{2} + 1|\)

এবং, \(\theta = 0\) হলে, \(\sec(0) = 1\) এবং \(\tan(0) = 0\) সুতরাং, \(\ln|\sec(0) + \tan(0)| = \ln|1 + 0| = \ln|1| = 0\)

অতএব, \(\left[ \ln|\sec(\theta) + \tan(\theta)| \right]_{0}^{\frac{\pi}{4}} = \ln(\sqrt{2} + 1) - 0 = \ln(\sqrt{2} + 1)\)

সুতরাং, \(\int_{0}^{\frac{\pi}{4}} \frac{\cos(\theta)}{\cos^2(\theta)} d\theta = \ln(\sqrt{2} + 1)\) 🤩