A) \[\frac{1}{2}log\text{(}sec\theta +tan\theta )+\frac{1}{2}sec\theta tan\theta +c\]
B) \[log(sec\theta +tan\theta )+sec\theta tan\theta +c\]
C) \[log(sec\theta -tan\theta )+\frac{1}{2}sec\theta tan\theta +c\]
D) None of the above
Correct Answer: A
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, \[\int{{{\sec }^{3}}\theta d\theta }=\frac{1}{2}\sec \theta \tan \theta \] \[+\frac{1}{2}\log (\sec \theta +\tan \theta )+c\] \[\left( \because {{c}_{1}}=\frac{c}{2} \right)\]
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