int_0^(π/4)secθdθ =?  

প্রশ্নঃ \(\displaystyle \int_{0}^{\pi/4} \sec \theta \, d\theta = ?\)

উত্তরঃ \(\displaystyle \ln(\sqrt{2} + 1)\)

সমাধান:

1. প্রথমে, আমরা জানি যে, \(\displaystyle \int \sec \theta \, d\theta = \ln | \sec \theta + \tan \theta | + C \)
2. তাই, নির্দিষ্ট সীমার জন্য, \[ \int_{0}^{\pi/4} \sec \theta \, d\theta = \left[ \ln | \sec \theta + \tan \theta | \right]_0^{\pi/4} \]
3. প্রথমে, \(\theta = \pi/4\) এ মান নির্ণয় করি: \[ \sec \left( \frac{\pi}{4} \right) = \frac{1}{\cos \left( \frac{\pi}{4} \right)} = \frac{1}{\frac{\sqrt{2}}{2}} = \sqrt{2} \] \[ \tan \left( \frac{\pi}{4} \right) = 1 \] অতএব, \[ \sec \left( \frac{\pi}{4} \right) + \tan \left( \frac{\pi}{4} \right) = \sqrt{2} + 1 \]
4. অতএব, \(\theta = 0\) এ মান: \[ \sec 0 = 1, \quad \tan 0 = 0 \] অর্থাৎ, \[ \sec 0 + \tan 0 = 1 + 0 = 1 \]
5. সুতরাং, সমাধান: \[ \int_{0}^{\pi/4} \sec \theta \, d\theta = \ln |\sqrt{2} + 1| - \ln |1| = \ln (\sqrt{2} + 1) - 0 = \ln (\sqrt{2} + 1) \]