A) \[{{\sin }^{2}}\left[ 1+\log \,\tan \frac{x}{2} \right]+c\]
B) \[\tan \left[ 1+\log \,\tan \frac{x}{2} \right]+c\]
C) \[{{\sec }^{2}}\left[ 1+\log \,\tan \frac{x}{2} \right]+c\]
D) \[-\tan \left[ 1+\log \,\tan \frac{x}{2} \right]+c\]
Correct Answer: B
Solution :
Let \[I=\int{\frac{\text{cosec x}}{{{\cos }^{2}}\left( 1+\log \tan \frac{x}{2} \right)}}dx\] Put \[1+\log \,\tan \frac{x}{2}=t\] \[\Rightarrow \] \[\frac{1}{\tan \frac{x}{2}}.{{\sec }^{2}}\frac{x}{2}.\frac{1}{2}dx=dt\] \[\Rightarrow \] \[\frac{1}{2\sin \frac{x}{2}\cos \frac{x}{2}}dx=dt\] \[\Rightarrow \] \[\text{cosec x dx = dt}\] \[\therefore \] \[I=\int{\frac{dt}{{{\cos }^{2}}t}}=\int{{{\sec }^{2}}}t\,\,dt\] \[=\tan \,\,t+c\] \[=\tan \,\,\left( 1+\log \,\,\tan \frac{x}{2} \right)+c\]
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