Question

Integrate the function in x sin^-1 x.

Answer 1

Let `I = int x sin^-1 x dx = int sin^-1 x* x dx`

`= sin^-1 x* (x^2/2) - int [d/dx (sin^-1 x) * x^2/2]  dx`

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

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

`= x^2/2 sin^-1 x - 1/2 I_1`

`I = x^2/2 sin^-1 x - 1/2 I_1`              ....(i)

Where `I_1 = int x^2/sqrt (1 - x^2)  dx`

Put x = sin θ 

⇒ dx = cosθ dθ

∴ `I_1 = int (sin^2 theta)/sqrt (1- sin^2 theta) cos d theta`

`= int (sin^2 theta)/(cos theta) * cos theta d theta`

`= int sin^2 theta d theta  = 1/2 int (1 - cos 2 theta) d theta`

`= 1/2int d theta - 1/2 int cos 2 theta d theta 1/2 theta - 1/2 (sin 2 theta)/2 + C`

  `1/2 theta - 1/2 sin theta cos theta + C`

`1/2 sin^-1x - 1/2x sqrt(1 - x^2) + C`                 ....(ii)

`[∵ sin theta = x ⇒ cos theta = sqrt (1 - sin^2 theta) = sqrt (1 - x^2)]`

From (i) and (ii), we get

∴ `I = x^2/2 sin^-1 x - 1/2 [1/2 sin^-1 x - 1/2 x sqrt(1 - x^2)] + C`

`= 1/4 sin^-1 x* (2x^2 - 1) + (x sqrt (1 - x^2))/4 + C`