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