The value of ∫03π2 tan−1⁡tan⁡x−sin−1⁡sin⁡xtan−1⁡tan⁡x+sin−1⁡sin⁡xdx is equal to

A
$\frac{π}{2}$
B
$π$
C
$\frac{3 π}{2}$
D
None of these

Solution:

$\begin{aligned} \int_{0}^{π / 2} \frac{|\tan^{- 1} \tan x| - |\sin^{- 1} \sin^{} x|}{|\tan^{- 1} \tan x| + |\sin^{- 1} \sin x|} dx + \int_{π / 2}^{3 π / 2} \frac{|\tan^{- 1} \tan x| - |\sin^{- 1} \sin^{} x|}{|\tan^{- 1} \tan x| + |\sin^{- 1} \sin^{- 1} x|} dx \end{aligned}$

Integrand, is discontinuous $\frac{π}{2}$ , then

$\begin{array}{l} \int_{0}^{π / 2} 0 \cdot dx + \int_{π / 2}^{3 π / 2} 0 \cdot dx = 0 \\ 0 < x < \frac{π}{2}, |\tan^{- 1} \tan x| = |\sin^{- 1} \sin x| \\ and \frac{π}{2} < x < \frac{3 π}{2}, |\tan^{- 1} \tan x| = |\sin^{- 1} \sin x| \end{array}$

The value of $\int_{0}^{\frac{3 π}{2}} \frac{|\tan^{- 1} \tan x| - |\sin^{- 1} \sin x|}{|\tan^{- 1} \tan x| + |\sin^{- 1} \sin x|} dx$ is equal to

Solution:

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