Mathematics

$$\int\limits_0^1 {{{\sin }^{ - 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right)dx} $$


SOLUTION
$$\begin{array} { l l } { \text { Let } x = \tan \theta } & { \Longrightarrow \theta = \tan ^ { - 1 } x } \\ { } & { \Longrightarrow \sin ^ { - 1 } \frac { 2 x } { 1 + x ^ { 2 } } = \sin ^ { - 1 } \frac { 2 \tan \theta } { 1 + \tan ^ { 2 } \theta } } \\ { } & { = \sin ^ { - 1 } ( \sin 2 \theta ) \quad \text { using } ( 1 ) } \\ { } & { = 2 \theta = 2 \tan ^ { - 1 } x } \end{array}$$
$$\begin{array} { l } { \text { Now the integral becomes fairly simple } } \\ { I = 2 \int _ { 1 } ^ { 0 } \tan ^ { - 1 } x \cdot 1 d x } \\ { \text { Integrating by parts } } \\ { \int u \cdot v d x = u \int v d x - \int \left( \frac { d u } { d x } \cdot \int v d x \right) d x } \end{array}$$
$$\begin{array} { l } { \text { Taking } v = 1 \quad \& \quad u = \tan ^ { - 1 } x } \\ { \Longrightarrow I = x \cdot \tan ^ { - 1 } \left. x \right| _ { 0 } ^ { 1 } - \left. \frac { \ln \left| 1 + x ^ { 2 } \right| } { 2 } \right| _ { 0 } ^ { 1 } } \end{array}$$
$$\Longrightarrow I =  \frac { \pi } { 4 } - \frac { \ln 2 } { 2 }$$
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Subjective Medium Published on 17th 09, 2020
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