How do you integrate ${sin}^{- 1} x d x$ $sin^-1x dx$ from 0 to 1?

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How do you integrate ${sin}^{- 1} x d x$ $sin^-1x dx$ from 0 to 1?

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Answer 1 · 2015-08-03T14:29:02

I would use $\int_{0}^{1} {sin}^{- 1} x$ $int_0^1 sin^-1x$ $d x = \frac{π}{2} - \int_{0}^{\frac{π}{2}} sin y$ $dx = pi/2 - int_0^(pi/2) siny$ $d y = \frac{π}{2} - 1$ $dy = pi/2 -1$ .

Explanation:

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

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

$y = {sin}^{- 1} x$ $y = sin^-1x" "$ $x = sin y$ $" "x = siny$

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

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

Finally notice that the area we are looking for is the part of the rectangle: $[0, 1] \times [0, \frac{π}{2}]$ $[0,1] xx [0, pi/2]$ that is not between the curve and the $y$ $y$ 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 $x = sin y$ $x = siny$ and right of $y$ $y$ axis from $y = 0$ $y=0$ to $y = \frac{π}{2}$ $y=pi/2$

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

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

$\int_{0}^{\frac{π}{2}} sin y d y = - cos y]_{0}^{\frac{π}{2}} = - 0 - (- 1) = 1$ $int_0^(pi/2) siny dy = -cosy]_0^(pi/2) = -0-(-1) = 1$

Therefore:

$\int_{0}^{1} {sin}^{- 1} x d x = \frac{π}{2} - 1$ $int_0^1 sin^-1 x dx = pi/2 - 1$

Answer 2 · 2015-08-03T15:41:05

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

Explanation:

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

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

so $sin t = x$ $sint = x$ and $d x = cos t$ $dx = cos t$ $d t$ $dt$

Substituting gets us:

$\int t cos t$ $int tcost$ $d t$ $dt$

We'll integrate by parts:

Let $u = t$ $u = t" "$ and $d v = cos t$ $" "dv = cost$ $d t$ $dt$

So $d u = d t$ $du = dt" "$ and $v = sin t$ $" "v = sint$

$u v - \int v d u = t sin t - \int sin t d t$ $uv-intvdu = tsint-intsint dt$

$= t sin t + cos t$ $= tsint + cost$

That is:

$\int t cos t$ $int tcost$ $d t$ $dt$ $= t sin t + cos t$ $= tsint + cost$

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

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

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

So
$\int_{0}^{1} {sin}^{- 1} x$ $int_0^1 sin^-1 x$ $d x$ $dx$ $= (x {sin}^{- 1} x + \sqrt{1 - x^{2}})}_{0}^{1}$ $= (x sin^-1x + sqrt(1-x^2))}_0^1$

$= (1 {sin}^{- 1} (1) + \sqrt{0}) - (0 {sin}^{- 1} (0) + \sqrt{1})$ $= (1 sin^-1(1) +sqrt0) - (0sin^-1(0) +sqrt(1))$

$= {sin}^{- 1} (1) - 1$ $= sin^-1(1) - 1$

$= \frac{π}{2} - 1$ $= pi/2 -1$

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