# How do you evaluate the integral int x^-2arcsinx?

Mar 25, 2017

$\int {x}^{-} 2 \arcsin x \mathrm{dx} = \ln \left\mid \frac{x}{\sqrt{1 - {x}^{2}}} \right\mid - \arcsin \frac{x}{x} + C$

#### Explanation:

Use integration by parts. Let:

$\left\{\begin{matrix}u = \arcsin x & \implies & \mathrm{du} = \frac{1}{\sqrt{1 - {x}^{2}}} \mathrm{dx} \\ \mathrm{dv} = {x}^{-} 2 \mathrm{dx} & \implies & v = - {x}^{-} 1\end{matrix}\right.$

Then:

$\int {x}^{-} 2 \arcsin x \mathrm{dx} = - \arcsin \frac{x}{x} + \int \frac{1}{x \sqrt{1 - {x}^{2}}} \mathrm{dx}$

Just working with the remaining integral, let $x = \sin \theta$. This implies that $\sqrt{1 - {x}^{2}} = \cos \theta$ and $\mathrm{dx} = \cos \theta d \theta$. Then:

$\int \frac{1}{x \sqrt{1 - {x}^{2}}} \mathrm{dx} = \int \frac{1}{\sin \theta \cos \theta} \cos \theta d \theta = \int \csc \theta d \theta$

Which is a commonly known integral. Also note that if $\sin \theta = x$, this is represented in a right triangle where the side opposite $\theta$ is $x$ and the hypotenuse is $1$. The leg adjacent to $\theta$, then, is $\sqrt{1 - {x}^{2}}$. From this right triangle we can say that $\csc \theta = \frac{1}{x}$ and $\cot \theta = \frac{\sqrt{1 - {x}^{2}}}{x}$, which will be relevant:

int1/(xsqrt(1-x^2))dx=-lnabs(csctheta+cottheta)=-lnabs(1/x+sqrt(1-x^2)/x

Combining the denominators and inverting the fraction by bringing the $- 1$ outside the natural log inside as a $- 1$ power, and putting this result into the original expression we found from integration by parts, we find a final answer of:

$\int {x}^{-} 2 \arcsin x \mathrm{dx} = \ln \left\mid \frac{x}{\sqrt{1 - {x}^{2}}} \right\mid - \arcsin \frac{x}{x} + C$

Mar 25, 2017

$\int {x}^{-} 2 \arcsin x \mathrm{dx} = \ln \left\mid \frac{x}{1 + \sqrt{1 - {x}^{2}}} \right\mid - \arcsin \frac{x}{x} + C$

#### Explanation:

Let $x = \sin \theta$. This implies that $\mathrm{dx} = \cos \theta d \theta$ and that $\theta = \arcsin x$. Then:

$I = \int {x}^{-} 2 \arcsin x \mathrm{dx} = \int {\csc}^{2} \theta \left(\theta\right) \left(\cos \theta d \theta\right) = \int \theta \cot \theta \csc \theta d \theta$

Then perform integration by parts. Let:

$\left\{\begin{matrix}u = \theta & \implies & \mathrm{du} = d \theta \\ \mathrm{dv} = \cot \theta \csc \theta d \theta & \implies & v = - \csc \theta\end{matrix}\right.$

So:

$I = u v - \int v \mathrm{du} = - \theta \csc \theta + \int \csc \theta d \theta$

Which is a common integral:

$I = - \theta \csc \theta - \ln \left\mid \csc \theta + \cot \theta \right\mid = \frac{- \theta}{\sin} \theta - \ln \left\mid \frac{1 + \cos \theta}{\sin} \theta \right\mid$

Rewriting the natural logarithm by bringing the $- 1$ into the integral as a $- 1$ power:

I=(-theta)/sintheta+lnabs(sintheta/(1+sqrt(1-sin^2theta))

And since $\sin \theta = x$:

$I = \ln \left\mid \frac{x}{1 + \sqrt{1 - {x}^{2}}} \right\mid - \arcsin \frac{x}{x} + C$