Integral-Calculus Question 503
Question: If $ I=\int{{{\sin }^{-\frac{11}{3}}}x{{\cos }^{-\frac{1}{3}}}xdx} $ $ =A{{\cot }^{2/3}}x+B{{\cot }^{8/3}}x+C $ . Then
Options:
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
Show Answer
Answer:
Correct Answer: B
Solution:
[b] If
$ I=\int{{{\sin }^{\frac{11}{3}}}}x{{\cos }^{\frac{1}{3}}}xdx, $ 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}}}( 1+{{\tan }^{2}}x ){{\sec }^{2}}xdx} $
Put $ \tan x=t,{{\sec }^{2}}xdx=dt $
$ I=\int{{t^{-\frac{11}{3}}}( 1+t^{2} )dt} $
$ =\int{( {t^{-\frac{11}{3}}}+{t^{-\frac{5}{3}}} )}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 $
BETA
