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प्रश्न
Integrate the function in `sin^(-1) ((2x)/(1+x^2))`.
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उत्तर
Let `I = sin^-1 ((2x)/ (1 + x^2)) dx`
Put x = tan t
⇒ dx = sec2 t dt
∴ `I = int sin^-1 ((2 tan t)/ (1 + tan^2 t)) sec^2 t dt`
`= int sin^-1 (sin 2t) sec^2 t dt`
`= 2t sec^2 t dt = 2 int sec^2 t dt`
`= 2 {t int sec^2 t dt - int [d/dt(t) * int sec^2 t dt] dt}`
`= 2 [t tant - int 1 * tan t dt]`
= 2 t tan t + 2 log |cos t| + C
`= 2 tan^-1 x*x + 2 log |1/ sqrt (1 + x^2)| + C` `...[∵ cos t = 1/ (sect) = 1/ (sqrt (1 + tan^2 t)) = 1/ (sqrt (1 + x^2))]`
`= 2 x tan^-1 x + 2 log |(1 + x^2)^(1/2)| + C`
`= 2 x tan^-1 x + 2 (- 1/2) log |1 + x^2| + C`
`= 2 x tan^-1 x - log |1 + x^2| + C`
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