A) \[\frac{-n({{w}_{2}}-{{w}_{1}})}{Rt}\]
B) \[({{\mu }_{s}}=1.4)\]
C) \[\frac{2}{\sqrt{3}}\left( \frac{\tau }{Bi} \right)\]
D) None of the above
Correct Answer: C
Solution :
Let\[I=\int_{{}}^{{}}{\frac{{{\sec }^{2}}x}{{{(\sec x+\tan x)}^{9/2}}}}dx\] Put sec x + tan x = t Then\[\sec x-\tan x=\frac{1}{t}\] and\[\sec x(\sec x+\tan x)dx=dt\] \[\Rightarrow \]\[\sec xdx=\frac{1}{t}dt\] Also,\[\sec x=\frac{1}{2}\left( t+\frac{1}{t} \right)\] \[\therefore \]\[I=\frac{1}{2}\int_{{}}^{{}}{\frac{\frac{1}{t}\left( t+\frac{1}{t} \right)}{{{t}^{9/2}}}}dt\] \[\Rightarrow \]\[I=\frac{1}{2}\int_{{}}^{{}}{\frac{1}{{{t}^{9/2}}}+\frac{1}{{{t}^{13/2}}}dt}\] \[\Rightarrow \]\[I=-\frac{1}{7{{t}^{7/2}}}-\frac{1}{11{{t}^{11/2}}}+K\] \[\Rightarrow \]\[I=-\frac{1}{{{t}^{11/2}}}\left\{ \frac{{{t}^{2}}}{7}+\frac{1}{11} \right\}+K\] \[\Rightarrow \]\[I=-\frac{1}{{{(\sec x+\tan x)}^{11/2}}}\] \[\left\{ \frac{1}{11}+\frac{1}{7}{{(\sec x+\tan x)}^{2}} \right\}+K\]
You need to login to perform this action.
You will be redirected in
3 sec
