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प्रश्न
Evaluate: `int_0^(π/2) sin 2x tan^-1 (sin x) dx`.
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उत्तर
`int_0^(π/2) sin 2x tan^-1 (sin x) dx`
= `int_0^(π/2) 2 sin x cos x tan^-1 (sin x)dx`
Let sin x = t
cos x dx = dt
When x = 0, t = 0 and x = `π/2`
`\implies` t = 1
`2int_0^1 t tan^-1 t dt`
= `2[tan^-1 t . t^2/2]_0^1 - 2int_0^1 1/(1 + t^2) . t^2/2 dt`
= `(π/4 . 1 - 0) - int_0^1 (t^2 + 1 - 1)/(1 + t^2)dt`
= `π/4 - int_0^1 (1 - 1/(1 + t^2))dt`
= `π/4 - [t - tan^-1 t]_0^1`
= `π/4 - [1 - π/4 - 0]`
= `π/4 - 1 + π/4`
= `π/2 - 1`.
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