Question

The integral `int sec^(2//3) "x cosec"^(4//3)"x dx"` is equal to (here C is a constant of integration)
A. `3tan^(-1//3)x+C`
B. `-3tan^(-1//3)x+C`
C. `-3cot^(-1//3)x+C`
D. `-(3)/(4)tan^(-4//3)x+C`

Answer 1

Correct Answer - B
Let `I=int sec^((2)/(3))x cos ec^((4)/(3))x dx = int(dx)/(cos^((2)/(3))x sin^((4)/(3))x) int(dx)/(((sin x)/(cos x))^((4)/(3))cos^((4)/(3)) x cos^((2)/(3))x)`
[dividing and multiplying by `cos^(4//3)` x in denominator]
`=int(dx)/(tan^((4)/(3)) x cos^(2) x)=int(sec^(2)xdx)/((tan x)^((4)/(3)))`
Now, put `tan x = t rArr sec^(2) x dx = dt`
`therefore I=int(dt)/(t^(4//3))=(t^((-4)/(3)+1))/((-4)/(3)+1)+C`
`=-3(1)/(t^((1)/(3)))+C =(-3)/((tan x)^((1)/(3)))+C=-3tan^(-(1)/(3))x+C`