JEE PYQ: Indefinite Integration Question 21
Question 21 - 2019 (09 Apr Shift 1)
The integral $\int \sec^{2/3}x\operatorname{cosec}^{4/3}x,dx$ is equal to:
(1) $-3\tan^{-1/3}x + C$
(2) $-\frac{3}{4}\tan^{-4/3}x + C$
(3) $-3\cot^{-1/3}x + C$
(4) $3\tan^{-1/3}x + C$
Show Answer
Answer: (1)
Solution
$I = \int \frac{\sec^2 x,dx}{\tan^{4/3}x}$. Put $\tan x = z$, $\sec^2 x,dx = dz$. $I = \int z^{-4/3},dz = \frac{z^{-1/3}}{-1/3} + C = -3(\tan x)^{-1/3} + C = -3\tan^{-1/3}x + C$.
BETA
