Question

निम्नलिखित प्रश्न को फलन का समाकलन कीजिए।

(sin-¹ x)²

Answer 1

माना `I = int (sin^-1 x)^2 . 1  dx`

(sin-¹ x)² को प्रथम व 1 को द्वितीय फलन लेने पर,

`= (sin^-1 x)^2 . x - int (2 sin^-1 x)/sqrt(1 - x^2). x  dx`

sin^-1 x = t प्रतिस्थापित करने पर,

`1/sqrt(1 - x^2)  dx = dt`

अत: `I = x (sin^-1 x)^2 - 2 int t sin t  dt`

`= x(sin^-1 x)^2 [-t cos t + sin t]`

`= x (sin^-1 x)^2 - 2 [-t cos t + int 1. sin t  dt]`

`= x (sin^-1 x)^2 + 2 t sqrt(1 - sin^2 t) - 2 sin t + C`

`= x(sin^-1  x)^2 + 2 sin^-1 x sqrt(1 - x^2) - 2x + C`

Answer 2

माना  `I = int (sin^-1 x)^2 dx`

`sin^-1 x = theta` रखने पर,

⇒ 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`