A) \[(1,\,1-\tan \theta )\]
B) \[(-1,\,1+\tan \theta )\]
C) \[(-1,\,1-\tan \theta )\]
D) \[(1,\,1+\tan \theta )\]
Correct Answer: B
Solution :
Let \[l=\int{\frac{{{\sec }^{2}}\theta d\theta }{\sec 2\theta +\tan 2\theta }}\,d\theta \] \[=\int{\frac{{{\sec }^{2}}\theta }{\frac{1+{{\tan }^{2}}\theta +2\tan \theta }{1-\tan 2\theta }}}\,d\theta \] \[=\int{\frac{{{\sec }^{2}}\theta (1-\tan \theta )}{1+\tan \theta }}\,d\theta \] Put \[\tan \theta =t\] \[\therefore \,\,\,\,\,\,\,\,{{\sec }^{2}}\theta d\theta =dt\] \[l=\int{\frac{\left( 1-t \right)dt}{1+t}}=\int{\left( -1+\frac{2}{1+t} \right)}dt=-t+2\log \left( 1+t \right)+c\] \[\therefore \,\,\,\,\,\,\,l=-\tan \theta +2{{\log }_{e}}|1+\tan \theta |+c\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\lambda =-1,\] \[f(\theta )=1+\tan \theta .\]
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