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Question
Evaluate the following : `int_0^1 sin^-1 ((2x)/(1 + x^2))*dx`
Solution
Let I = `int_0^1 sin^-1 ((2x)/(1 + x^2))*dx`
Put x = tan t, i.e. t = tan–1x
∴ dx = sec2t·dt
When x = 0, t = tan–1 0 = 0
When x = 1, t = tan–1 = `pi/(4)`
∴ I = `int_0^(pi/4) sin^-1 ((2tant)/(1 + tan^2t))sec^2t*dt`
= `int_0^(pi/4) sin^-1 (sin2t) sec^2t*dt`
= `int_0^(pi/4) 2t sec^2t*dt`
= `[2t int sec^2t*dt]_0^(pi/4) - int_0^(pi/4) [d/dx (2t) int sec^2t*dt]`
= `[2t tan t]_0^(pi/4) - int_0^(pi/4) 2 tan t *dt`
= `[2* pi/4 tan pi/4 - 0] - 2log(sec t)]_0^(pi/4)`
= `pi/2 - 2[log(sec pi/4) - log (sec0)]`
= `pi/2 - 2[log sqrt(2) - log 1]`
= `pi/2 - 2[1/2 log 2 - 0]`
= `pi/(2) - log 2`.
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