Mathematics

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


ANSWER

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
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