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Q. The value of $ \int{{{\sec }^{3}}\theta d\theta } $ is

Rajasthan PETRajasthan PET 2004

Solution:

Let $ I=\int{{{\sec }^{3}}\theta }d\theta $
$ =\int{\underset{I}{\mathop{\sec \theta }}\,\underset{II}{\mathop{{{\sec }^{2}}\theta }}\,d\theta } $
$ =\sec \theta (\int{{{\sec }^{2}}\theta d\theta )} $
$ -\int{\left[ \left\{ \frac{d}{dx}(\sec \theta ) \right\}\int{{{\sec }^{2}}\theta }d\theta \right]}\,d\theta $
$ =\sec \theta .\tan \theta -\int{\sec \theta \tan \theta .\tan \theta d\theta } $
$ =\sec \theta .\tan \theta -\int{\sec \theta {{\tan }^{2}}\theta .d\theta } $
$ \Rightarrow $ $ I=\sec \theta \tan \theta -\int{\sec \theta ({{\sec }^{2}}\theta -1)d\theta } $
$ \Rightarrow $ $ I=\sec \theta \tan \theta -\int{{{\sec }^{3}}\theta d\theta +\int{\sec \theta }d\theta } $
$ \Rightarrow $ $ I=\sec \theta \tan \theta -I+\log (\sec \theta +\tan \theta )+{{c}_{1}} $
$ \Rightarrow $ $ 2I=\sec \theta \tan \theta +\log (\sec \theta +\tan \theta )+{{c}_{1}} $
$ \Rightarrow $ $ I=\frac{1}{2}\sec \theta \tan \theta +\frac{1}{2}\log (\sec \theta +\tan \theta )+\frac{{{c}_{1}}}{2} $
Hence, $ \int{{{\sec }^{3}}\theta d\theta }=\frac{1}{2}\sec \theta \tan \theta $
$ +\frac{1}{2}\log (\sec \theta +\tan \theta )+c $
$ \left( \because {{c}_{1}}=\frac{c}{2} \right) $