# Question dfe07

Jan 14, 2018

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

#### Explanation:

Given: $\int \arcsin \frac{x}{\sqrt{1 - {x}^{2}}} \mathrm{dx}$

Apply u-substitution

Let $u = \arcsin \left(x\right)$

$\mathrm{du} = \frac{1}{\sqrt{1 - {x}^{2}}} \mathrm{dx}$

See Proof Below

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Let: $y = \arcsin \left(x\right)$

Take the $\sin$ of both sides. So

$\sin \left(y\right) = x$

Use implicit differentation to differentiate both sides

$\frac{\mathrm{dy}}{\mathrm{dx}} \cdot \cos \left(y\right) = 1$

Divde by $\cos \left(y\right)$ on both sides

dy/dx=1/color(red)(cos(y)

Rewrite in terms of $x$

Since $\sin \left(y\right) = \frac{x}{1}$

Then color(red)(cos(y)=sqrt(1^2-x^2)/1=sqrt(1-x^2)

So rewritng we get

dy/dx=1/color(red)(cos(y))=color(red)(1/sqrt(1-x^2)#

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Thus, the integral now becomes:

$\int u \mathrm{du}$

Since $\int {x}^{a} = {x}^{a + 1} / \left(a + 1\right)$, we solve the integral to get:

$= {u}^{1 + 1} / \left(1 + 1\right) = {u}^{2} / 2$

Reverse the subsitution:

$= {\arcsin}^{2} \frac{x}{2} + \text{C}$