The integral $\int \sec ^{2 / 3} x \operatorname{cosec}^{4 / 3} x d x$ is equal to: (Here $C$ is a constant of integration)

(B) $-\frac{3}{4} \tan ^{-4 / 3} x+C$

JEE Main · 2019 · 09 Apr 2019 · Shift 1 · Indefinite Integrals

The integral

\int sec^{2/3} x cosec^{4/3} x d x

is equal to: (Here

C

is a constant of integration)

Answer: (D)

Rewrite the integrand as

\frac{1}{s i n ^{4/3} x c o s ^{2/3} x} = \frac{s e c ^{2} x}{( t a n x ) ^{4/3}}

. With

u = tan x

d u = sec^{2} x d x

, it becomes

\int u^{- 4/3} d u = \frac{u ^{- 1/3}}{- 1/3} = - 3 u^{- 1/3}

, that is

- 3 tan^{- 1/3} x + C

JEE Main · 2019 · 09 Apr 2019 · Shift 1 · Single correct option

If …
JEE Main 2026View solution →
Let $f (x) = \int \frac{( 2 - x ^{2} ) \cdot e ^{x}}{( 1 + x ) ( 1 - x ) ^{3/2}} d x$ . If $f (0) = 0$ , then $f (\frac{1}{2})$ is equal to :
JEE Main 2026View solution →
Let $I (x) = \int \frac{3 dx}{( 4 x + 6 ) ( 4 x ^{2} + 8 x + 3 )}$ and $I (0) = \frac{3}{4} + 20$ . If $I (\frac{1}{2}) = \frac{a 2}{b} + c$ , where …
JEE Main 2026View solution →
Let $f (t) = \int (\frac{1 - s i n ( l o g _{e} t )}{1 - c o s ( l o g _{e} t )}) dt, t > 1$ .

If $f (e^{π /2}) = - e^{π /2}$ and …
JEE Main 2026View solution →
Let $f (x) = \int \frac{7 x ^{10} + 9 x ^{8}}{( 1 + x ^{2} + 2 x ^{9} ) ^{2}} d x, x > 0$ ,
$lim_{x \to 0} f (x) = 0$ and $f (1) = \frac{1}{4}$ . …
JEE Main 2026View solution →
If $\int (\frac{1 - 5 c o s ^{2} x}{s i n ^{5} x c o s ^{2} x}) d x = f (x) + C$ where $C$ is the constant of integration, then $f (\frac{π}{6}) - f (\frac{π}{4})$ is equal to
JEE Main 2026View solution →