Evaluate the definite integral _0^ 2 2 x ^ -1 ( x) d x

Question

Evaluate the definite integral $\int_0^{\frac{\pi}{2}} \sin 2 x \tan ^{-1}(\sin x) d x$

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Answer

According to the question , $I =\int _ { 0 } ^ { \pi / 2 } 2 \sin x \cos x \tan ^ { - 1 } ( \sin x ) d x$
Put , sin x = t
$\Rightarrow$ cos x dx = dt
Lower Limit , when x= 0, then t = sin 0 =0
Upper Limit , when x =$\frac{\pi}{2}$ , then t = sin$\frac{\pi}{2}$ = 1.
$\therefore \quad I = 2 \int _ { 0 } ^ { 1 } t \times \tan ^ { - 1 } t d t$
Applying integration by parts,we get
$I = 2 \left[ \frac { t ^ { 2 } } { 2 } \times \tan ^ { - 1 } t \right] _ { 0 } ^ { 1 } - 2 \int _ { 0 } ^ { 1 } \frac { 1 } { 1 + t ^ { 2 } } \times \frac { t ^ { 2 } } { 2 } d t$
$= 2 \left[ \frac { t ^ { 2 } } { 2 } \times \tan ^ { - 1 } t \right] _ { 0 } ^ { 1 } - \int _ { 0 } ^ { 1 } \frac { t ^ { 2 } } { 1 + t ^ { 2 } } d t$
$= 2 \times \frac { 1 } { 2 } \times \tan ^ { - 1 } ( 1 ) - \int _ { 0 } ^ { 1 } \frac { 1 + t ^ { 2 } - 1 } { 1 + t ^ { 2 } } d t$
$= 1 \times \frac { \pi } { 4 } - \int _ { 0 } ^ { 1 } \left( \frac { 1 + t ^ { 2 } } { 1 + t ^ { 2 } } - \frac { 1 } { 1 + t ^ { 2 } } \right) d t$
$= \frac { \pi } { 4 } - \int _ { 0 } ^ { 1 } \left( 1 - \frac { 1 } { 1 + t ^ { 2 } } \right) d t$
$= \frac { \pi } { 4 } - \left[ t - \tan ^ { - 1 } t \right] _ { 0 } ^ { 1 }$
$= \frac { \pi } { 4 } - 1 + \tan ^ { - 1 } 1$
$= \frac { \pi } { 4 } - 1 + \frac { \pi } { 4 } = \frac { 2 \pi } { 4 } - 1$
$ = \frac { \pi } { 2 } - 1 $
$\therefore \quad I = \frac { \pi } { 2 } - 1 $

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