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

Aug 3, 2015

I would use ${\int}_{0}^{1} {\sin}^{-} 1 x$ $\mathrm{dx} = \frac{\pi}{2} - {\int}_{0}^{\frac{\pi}{2}} \sin y$ $\mathrm{dy} = \frac{\pi}{2} - 1$.

#### Explanation:

We want the area under $y = {\sin}^{-} 1 x$ and above $\left[0 , 1\right]$

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 ${\sin}^{-} 1 x$ and use that, the geometry may provide a simpler solution. (Although the explanation will take a few lines.)

$y = {\sin}^{-} 1 x \text{ }$ $\text{ } x = \sin y$

For $x$ in the interval $\left[0 , 1\right]$, the $y$ values (the range) include everything from $0$ to $\frac{\pi}{2}$.

So, the curve can be described by $x = \sin y$ for $y \in \left[0 , \frac{\pi}{2}\right]$

Finally notice that the area we are looking for is the part of the rectangle: $\left[0 , 1\right] \times \left[0 , \frac{\pi}{2}\right]$ 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 $x = \sin y$ and right of $y$ axis from $y = 0$ to $y = \frac{\pi}{2}$

Area of rectangle = base x height $= 1 \times \frac{\pi}{2} = \frac{\pi}{2}$

Area left of $x = \sin y$ and right of $y$ axis from $y = 0$ to $y = \frac{\pi}{2}$ is:

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

Therefore:

${\int}_{0}^{1} {\sin}^{-} 1 x \mathrm{dx} = \frac{\pi}{2} - 1$

Aug 3, 2015

Here it is using the indefinite integral of ${\sin}^{-} 1 x \mathrm{dx}$.

#### Explanation:

$\int {\sin}^{-} 1 x$ $\mathrm{dx}$

Let $t = {\sin}^{-} 1 x$,

so $\sin t = x$ and $\mathrm{dx} = \cos t$ $\mathrm{dt}$

Substituting gets us:

$\int t \cos t$ $\mathrm{dt}$

We'll integrate by parts:

Let $u = t \text{ }$and $\text{ } \mathrm{dv} = \cos t$ $\mathrm{dt}$

So $\mathrm{du} = \mathrm{dt} \text{ }$ and $\text{ } v = \sin t$

$u v - \int v \mathrm{du} = t \sin t - \int \sin t \mathrm{dt}$

$= t \sin t + \cos t$

That is:

$\int t \cos t$ $\mathrm{dt}$$= t \sin t + \cos t$

With $t = {\sin}^{-} 1 x$, and $\sin t = x$, we also have

$\cos t = \sqrt{1 - {\sin}^{2} t} = \sqrt{1 - {x}^{2}}$, so

$\int {\sin}^{-} 1 x$ $\mathrm{dx}$ $= x {\sin}^{-} 1 x + \sqrt{1 - {x}^{2}} + C$

So
${\int}_{0}^{1} {\sin}^{-} 1 x$ $\mathrm{dx}$  = (x sin^-1x + sqrt(1-x^2))}_0^1

$= \left(1 {\sin}^{-} 1 \left(1\right) + \sqrt{0}\right) - \left(0 {\sin}^{-} 1 \left(0\right) + \sqrt{1}\right)$

$= {\sin}^{-} 1 \left(1\right) - 1$

$= \frac{\pi}{2} - 1$