Question

How do you integrate `sin^-1x dx` from 0 to 1?

Answer 1

I would use `int_0^1 sin^-1x` `dx = pi/2 - int_0^(pi/2) siny` `dy = pi/2 -1`.

Explanation:

We want the area under `y = sin^-1 x` and above `[0,1]`

graph{(y - arcsinx)(y) sqrt(x-x^2)/sqrt(x-x^2) <= 0 [-1.207, 3.124, -0.284, 1.875]}

While we could find the indefinite integral of `sin^-1x` and use that, the geometry may provide a simpler solution. (Although the explanation will take a few lines.)

`y = sin^-1x" "` `" "x = siny`

For `x` in the interval `[0,1]`, the `y` values (the range) include everything from `0` to `pi/2`.

So, the curve can be described by `x = siny` for `y in [0, pi/2]`

Finally notice that the area we are looking for is the part of the rectangle: `[0,1] xx [0, pi/2]` that is not between the curve and the `y` axis.

We want the area of the rectangle minus the area in blue below:

graph{(y - arcsinx)(y-(pi/2)) sqrt(x-x^2)/sqrt(x-x^2) <= 0 [-1.083, 2.767, -0.078, 1.841]}

Area of rectangle minus area left of `x = siny` and right of `y` axis from `y=0` to `y=pi/2`

Area of rectangle = base x height `= 1xxpi/2 = pi/2`

Area left of `x = siny` and right of `y` axis from `y=0` to `y=pi/2` is:

`int_0^(pi/2) siny dy = -cosy]_0^(pi/2) = -0-(-1) = 1`

Therefore:

`int_0^1 sin^-1 x dx = pi/2 - 1`

Answer 2

Here it is using the indefinite integral of `sin^-1 x dx`.

Explanation:

`int sin^-1x` `dx`

Let `t = sin^-1 x`,

so `sint = x` and `dx = cos t` `dt`

Substituting gets us:

`int tcost` `dt`

We'll integrate by parts:

Let `u = t" "`and `" "dv = cost` `dt`

So `du = dt" "` and `" "v = sint`

`uv-intvdu = tsint-intsint dt`

`= tsint + cost`

That is:

`int tcost` `dt` `= tsint + cost`

With `t = sin^-1 x`, and `sint = x`, we also have

`cost = sqrt(1-sin^2t) = sqrt(1-x^2)`, so

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

So
`int_0^1 sin^-1 x` `dx` `= (x sin^-1x + sqrt(1-x^2))}_0^1`

`= (1 sin^-1(1) +sqrt0) - (0sin^-1(0) +sqrt(1))`

`= sin^-1(1) - 1`

`= pi/2 -1`