If $\int \sqrt{\sec 2x-1} dx=\alpha \log_{e}\left|\cos 2x+\beta+\sqrt{\cos 2x\left(1+\cos 2x\right)}\right|$ + constant, then $\beta-\alpha$ is equal to ____.

JEE Main · 2023 · 30 Jan 2023 · Shift 2 · Indefinite Integrals

\int sec 2 x - 1 d x = α lo g_{e} cos 2 x + β + cos 2 x (1 + cos 2 x)

+ constant, then

β - α

is equal to ____.

Answer: 1

Since

sec 2 x - 1 = \frac{1 - cos 2 x}{cos 2 x} = \frac{2 sin x}{cos 2 x}

and

cos 2 x = 2 cos^{2} x - 1

, the substitution

t = cos x

gives

- 2 \int \frac{d t}{2 t ^{2} - 1} = - ln 2 cos x + cos 2 x + C

. Writing this as half the logarithm of the square,

- \frac{1}{2} ln 2 cos^{2} x + cos 2 x + 22 cos x cos 2 x = - \frac{1}{2} ln cos 2 x + \frac{1}{2} + cos 2 x (1 + cos 2 x) + C

. Comparing with the given form,

α = - \frac{1}{2}

and

β = \frac{1}{2}

, so

β - α = 1

JEE Main · 2023 · 30 Jan 2023 · Shift 2 · Integer answer

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 →