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\[l=\int{\frac{\text{cosec}\,x}{{{\cos }^{2}}\left( 1+\log \tan \frac{x}{2} \right)}dx}\] Put\[\left( 1+\log \tan \frac{x}{2} \right)=t\Rightarrow \frac{1}{\tan \frac{x}{2}}{{\sec }^{2}}\frac{x}{2}\cdot \frac{1}{2}dx=dt\] \[\Rightarrow \] \[\text{cosec}\,x\,\,dx=dt\] \[\therefore \] \[l=\int{\frac{1}{{{\cos }^{2}}t}dt}=\int{{{\sec }^{2}}t\,\,dt=\tan t+C}\] \[=\tan \left[ 1+\log \tan \frac{x}{2} \right]+C\]
You need to login to perform this action.
You will be redirected in
3 sec
