Mathematics

# The value of $\displaystyle \int_{-2\pi}^{5\pi} \cot^{-1}(\tan x) dx$ is equal to:

None of these

##### SOLUTION

Let $\cot^{-1}(\tan x) = \theta$

Then, $\tan x = \cot \theta$

Differentiating,

$\sec^2 x \; dx = -\csc ^2 \theta \; d\theta$

$dx = -\dfrac{\csc ^2 \theta }{\sec^2 x}\; d\theta$

Since

$\tan x = \cot \theta$

$\Rightarrow \pm \sqrt{\sec^2 x - 1} = \cot \theta$

$\sec^2 x - 1 = \cot^2 \theta$

$\sec^2 x = 1 + \cot^2 \theta = \csc^2 \theta$

Hence,

$dx = -\dfrac{\csc ^2 \theta }{\sec^2 x}\; d\theta = -d\theta$

$\therefore \int \cot^{-1}(\tan x) dx= -\int \theta d\theta$

$= -\dfrac{\theta^2}{2} + c$

Since the function $y = \cot^{-1}x$ has the range $(0,\pi)$,

$\cot^{-1}(\tan (-2\pi)) = \cot^{-1}(0) = \dfrac{\pi}{2}$

$\cot^{-1}(\tan (5\pi)) = \cot^{-1}(0) = \dfrac{\pi}{2}$

$\therefore \displaystyle \int_{-2\pi}^{5\pi} \cot^{-1}(\tan x) dx= -\dfrac{\theta^2}{2} \bigg |_{\frac{\pi}{2}}^{\frac{\pi}{2}} = 0$

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Single Correct Hard Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 114

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