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प्रश्न
Evaluate: `int_0^(pi/2) sin2x*tan^-1 (sinx)*dx`
उत्तर
Let I = `int_0^(pi/2) sin2x*tan^-1 (sinx)*dx`
= `int_0^(pi/2) tan^-1 (sinx)*(2sinx cosx)*dx`
Put sinx = t
∴ cos x·dx = dt
When x = 0, t = sin 0 = 0
When x = `pi/(2), t = sin pi/(2)` = 1
∴ I = `int_0^1(tan^-1 t)(2t)*dt`
= `[tan^-1 t int 2t dt]_0^1 - int_0^1(d/dt (tan^-1 t) int 2t dt)*dt`
= `[tan^-1 t int (t)^2 ]_0^1 - int_0^1 1/(1 + t^2)*t^2*dt`
= `[t^2 tan^-1 t]_0^1 - int_0^1 ((1 + t^2) - 1)/(1 + t^2)*dt`
= `[1*tan^-1 - 0] -int_0^1 (1 - 1/(1 + t^2))*dt`
= `pi/(4) - [t - tan^-1 t]_0^1`
= `pi/(4) - [(1 - tan^-1 1) - 0]`
= `pi/(4) - 1 + pi/(4)`
= `pi/(2) - 1`
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