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प्रश्न
Evaluate the definite integral:
`int_0^(pi/2) sin 2x tan^(-1) (sinx) dx`
उत्तर
Let I = `int_0^(pi/2) sin 2x tan^-1 (sin x) dx`
`= 2 int_0^(pi/2) sin x cos x tan^-1 (sin x) dx`
Putting sin x = t, cos x dx = dt
When x = 0, t = sin 0 ⇒ t = 0
And when `x = pi/2, t = sin pi/2`
=> t = 1
∴ `I = 2 int_0^1 t tan^-1 t dt`
`= 2 [tan^-1 (t) t^2/2]_0^1 - 2 int_0^1 1/ (1 + t^2)* t^2/2 dt`
`= 2 [t^2/2 tan^-1 (t)]_0^1 - 2/2 int_0^1 (1 + t^2 - 1)/ (1 + t^2) dt`
`= [t^2 tan^-1 (t)]_0^1 - int_0^1 (1 - 1/ (1 + t^2)) dt`
`= [t^2 tan^-1 (t) - t + tan^-1 t]_0^1`
`= tan^-1 (1) - 1 + tan^-1`
`= pi/4 - 1 + pi/4`
`= pi/2 - 1`
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