A) \[{{[{{\log }_{e}}\,(tan\,x)]}^{2}}+c\]
B) \[\frac{1}{2}{{({{\log }_{e}}\,\tan x)}^{2}}+c\]
C) \[{{\log }_{e}}\,\,({{\log }_{e}}\,\,\tan \,x)+c\]
D) \[{{\log }_{e}}\,\,\tan \,x+c\]
Correct Answer: B
Solution :
\[\int{\frac{{{\log }_{e}}\,(\tan x)}{\sin \,x\,\cos \,x}}\,dx=\int{\frac{{{\log }_{e}}\,(\tan x)}{\tan \,x}}\,{{\sec }^{2}}\,x\,dx\] Put \[\tan \,x=t\Rightarrow se{{c}^{2}}\,dx=dt\] \[\therefore \] \[\int{\frac{{{\log }_{e}}\,(\tan x)}{\sin \,x\,\cos \,x}}\,dx=\int{\frac{{{\log }_{e}}t}{t}}\,dt\] Again, put \[{{\log }_{e}}t=u\] \[\Rightarrow \] \[\frac{1}{t}\,dt=du\] \[\therefore \] \[\int{\frac{{{\log }_{e}}\,(\tan x)}{\sin x\,\cos \,x}}\,dx=\int{u\,\,du}\] \[=\frac{{{u}^{2}}}{2}+c\] \[=\frac{{{({{\log }_{e}}t)}^{2}}}{2}+c\] \[=\frac{1}{2}{{[{{\log }_{e}}\,(\tan x)]}^{2}}+c\]
You need to login to perform this action.
You will be redirected in
3 sec
