If $I = 3 π \int 0 [tan x] d x,$ where $[.]$ represents the greatest integer function, then the value of $∣ ∣ ∣ \frac{2 I}{π} ∣ ∣ ∣$ is

Open in App

Solution

$I = 3 π \int 0 [tan x] d x \dots (1)$
Also, $I = 3 π \int 0 [tan (3 π - x)] d x$
$\Rightarrow I = 3 π \int 0 [- tan x] d x \dots (2)$
Adding $(1)$ and $(2)$ ,
$2 I = 3 π \int 0 ([tan x] + [- tan x]) d x$
$\Rightarrow 2 I = 3 π \int 0 - d x (∵ [x] + [- x] = - 1 \forall x \notin Z)$
$\Rightarrow I = - \frac{3 π}{2}$

$\Rightarrow ∣ ∣ ∣ \frac{2 I}{π} ∣ ∣ ∣ = 3$

Suggest Corrections

Similar questions

I = 3 π \int 0 [tan x] d x,

[.]

∣ ∣ ∣ \frac{2 I}{π} ∣ ∣ ∣

\int_{0}^{3 π / 2} sin [\frac{2 x}{π}] \cdot d x

[\cdot]

cos x = \frac{2 cos y - 1}{2 - cos y}

x, y \in (0, π)

tan \frac{x}{2} cot \frac{y}{2} = k

[k]

\frac{6}{π} π / 3 \int π / 6 [2 sin x] d x

f (x) = \int \frac{d x}{tan x + sec x + cot x + cosec x}

f (0) = \frac{5}{2},

[f (π)]

Join BYJU'S Learning Program