If $\int_{0}^{\frac{\pi}{3}} \frac{\tan \theta}{\sqrt{2 \mathrm{k} \sec \theta}} \mathrm{d} \theta=1-\frac{1}{\sqrt{2}},(\mathrm{k}>0)$, then the value of k is:

JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Definite Integrals

\int_{0}^{\frac{π}{3}} \frac{t a n θ}{2 k s e c θ} d θ = 1 - \frac{1}{2}, (k > 0)

, then the value of k is:

Answer: (A)

Pull the constant out:

\int_{0}^{π /3} \frac{t a n θ}{2 k s e c θ} d θ = \frac{1}{2 k} \int_{0}^{π /3} \frac{s i n θ}{c o s θ} d θ

. Substitute

u = cos θ

d u = - sin θ d θ

, with

u

running from

1

\frac{1}{2}

, giving

\frac{1}{2 k} \int_{1/2}^{1} u^{- 1/2} d u = \frac{1}{2 k} [2 u]_{1/2}^{1} = \frac{2 - 2}{2 k}

. Writing

2 - 2 = 2 (2 - 1)

and

2 k = 2 k

, this is

\frac{2 - 1}{k}

. Setting it equal to

1 - \frac{1}{2} = \frac{2 - 1}{2}

and cancelling

2 - 1

leaves

k = 2

, so

k = 2

JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Single correct option

The value of $\int_{- π /6}^{π /6} (\frac{π + 4 x ^{11}}{1 - s i n ( ∣ x ∣ + π /6 )}) d x$ is equal to
JEE Main 2026View solution →
$6 \int_{0}^{π} ∣ (sin 3 x + sin 2 x + sin x) ∣ d x$ is equal to.......
JEE Main 2026View solution →
Let $[\cdot]$ denote the greatest integer function and $f (x) = lim_{n \to \infty} \frac{1}{n ^{3}} \sum_{k = 1}^{n} [\frac{k ^{2}}{3 ^{x}}]$ . Then $12 \sum_{j = 1}^{\infty} f (j)$ is equal …
JEE Main 2026View solution →
If $\int_{0}^{1} 4 cot^{- 1} (1 - 2 x + 4 x^{2}) dx = a tan^{- 1} (2) - b lo g_{e} (5)$ , where $a, b \in N$ , then $(2 a + b)$ is equal to ____.
JEE Main 2026View solution →
Let $f : [1, \infty) \to R$ be a differentiable function. If $6 \int_{1}^{x} f (t) d t = 3 x f (x) + x^{3} - 4$ for all $x \geq 1$ , then the value of $f (2) - f (3)$ is
JEE Main 2026View solution →
The value of $\int_{- \frac{π}{2}}^{\frac{π}{2}} (\frac{1}{[ x ] + 4}) dx$ , where [•] denotes the greatest integer function, is
JEE Main 2026View solution →