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प्रश्न
Integrate the function in (sin-1x)2.
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उत्तर
Let `I = int (sin^-1 x)^2 dx`
Put `sin^-1 x = theta`
⇒ x = sinθ
⇒ dx = cosθ dθ
∴ `I = int theta^2 cos theta d theta`
`= theta^2 int (cos theta) d theta - int (d/ (d theta) (theta^2) * int cos theta d theta) d theta`
`= theta^2 (sin theta) - int 2 theta (sin theta) d theta`
`= theta^2 sin theta - 2 int theta sin theta d theta + C`
`= theta^2 sin theta - 2 [theta * (- cos theta) - int 1 * (- cos theta) d theta] + C`
`= theta^2 sin theta + 2 theta cos theta - 2 int cos theta d theta + C`
`= theta^2 sin theta + 2 theta sqrt (1 - sin^2 theta) - 2 sin theta + C`
`= x (sin^-1 x)^2 + 2sin^-1 x sqrt (1 - x^2) - 2x + C`
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