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Question
Evaluate : `int_0^1 "x" . "tan"^-1 "x" "dx"`
Solution
Let I = `int_0^1 "x" . "tan"^-1 "x" "dx"`
Integrating by parls, we get
I = `["tan"^-1 "x" . int "x" "dx"]_0^1 - int_0^1 ("d"/"dx" . "tan"^-1 "x") . int "x dx"`
`= ["tan"^-1 "x" . "x"^2/2]_0^1 - int _0^1 1/(1 + "x"^2) . "x"^2/2 "dx"`
`= ["tan"^-1 "x" . "x"^2/2]_0^1 - 1/2 int _0^1 "x"^2/("x"^2 + 1) "dx"`
`= ["tan"^-1 "x" . "x"^2/2]_0^1 - 1/2 int _0^1 (("x"^2 + 1)-1)/("x"^2 + 1) "dx"`
`= ["tan"^-1 (1) . 1/2 - 0] - 1/2 [int _0^1 1 "dx" - int_0^1 1/(1+"x"^2) "dx"]`
`= [pi/8] - 1/2 ["x" - "tan"^-1 "x"]_0^1`
`= pi/8 - 1/2 [1 - "tan"^-1 (1) - (0 - 0)]`
`= pi/8 -1/2 [1 - pi/4] = pi/8 - 1/2 + pi/8 = pi/4 -1/2`
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