A) \[A=\frac{2}{3},B=\frac{8}{3}\]
B) \[A=-\frac{3}{2},B=-\frac{3}{8}\]
C) \[A=\frac{3}{2},B=\frac{3}{8}\]
D) None of these
Correct Answer: B
Solution :
| [b] If |
| \[I=\int{{{\sin }^{\frac{11}{3}}}}x{{\cos }^{\frac{1}{3}}}x\,\,dx,\] here \[\frac{-\frac{11}{3}-\frac{1}{3}-2}{2}=-3\] |
| (a negative integer) |
| \[I=\int{\frac{{{\sin }^{-\frac{11}{3}}}x}{{{\cos }^{-\frac{11}{3}}}}{{\cos }^{-\frac{1}{3}}}x.{{\cos }^{-\frac{11}{3}}}xdx}\] |
| \[=\int{{{(\tan x)}^{-\frac{11}{3}}}{{(\cos \,\,x)}^{-4}}dx}\] |
| \[=\int{{{(\tan x)}^{-\frac{11}{3}}}x.{{\sec }^{4}}xdx}\] |
| \[=\int{{{(\tan x)}^{-\frac{11}{3}}}\left( 1+{{\tan }^{2}}x \right){{\sec }^{2}}xdx}\] |
| Put \[\tan x=t,{{\sec }^{2}}xdx=dt\] |
| \[I=\int{{{t}^{-\frac{11}{3}}}\left( 1+{{t}^{2}} \right)dt}\] |
| \[=\int{\left( {{t}^{-\frac{11}{3}}}+{{t}^{-\frac{5}{3}}} \right)}dt=\frac{{{t}^{-\frac{8}{3}}}}{-\frac{8}{3}}+\frac{{{t}^{-\frac{2}{3}}}}{-\frac{2}{3}}+C\] |
| \[=-\frac{3}{8}{{(\tan x)}^{-\frac{8}{3}}}-\frac{3}{2}{{(\tan x)}^{-\frac{2}{3}}}+C\] |
| \[=-\frac{3}{2}{{\cot }^{\frac{2}{3}}}x-\frac{3}{8}{{\cot }^{8/3}}x+C\] |
You need to login to perform this action.
You will be redirected in
3 sec
