Integate with arcsin x^1/2

**Revy** · August 23rd 2010, 08:45 AM

I need to integrate with direct integration method
$\int\frac{\arcsin \sqrt x}{\sqrt{x - x^2}}$

I do get $d(x-x^2)=1-2x d(x-x^2)$ then $\int\frac{\arcsin \sqrt x}{(x - x^2)^\frac{1}{2}}\frac{1}{1-2x}\d(x-x^2)=$ and that where my brains stop, I don't know what to to with that $\arcsin \sqrt x$

**Opalg** · August 23rd 2010, 10:55 AM

Originally Posted by Revy

I need to integrate with direct integration method
$\int\frac{\arcsin \sqrt x}{\sqrt{x - x^2}}$

I do get $\d(x-x^2)=1-2x\d(x-x^2)$ then $\int\frac{\arcsin \sqrt x}{(x - x^2)^\frac{1}{2}}\frac{1}{1-2x}\d(x-x^2)=$ and that where my brains stop, I don't know what to to with that $\arcsin \sqrt x$

Try making the substitution $u = \arcsin\sqrt x$ . Then $\frac{du}{dx} = \frac1{2\sqrt{1-x}\sqrt x} = \frac1{2\sqrt{x-x^2}}$ and the integral becomes $\displaystyle\int\!\!2u\,du = u^2 = \bigl(\arcsin\sqrt x\bigr)^2$ (plus a constant of integration).

**Revy** · August 23rd 2010, 11:23 AM

For this task I can't use substitution, have to do it directly. But thanks, this one will be useful as well

Just one thing, how do you get denominator in
$\frac{du}{dx} = \frac1{2\sqrt{1-x}\sqrt x} = \frac1{2\sqrt{x-x^2}}$ ?

**BayernMunich** · August 23rd 2010, 05:53 PM

Originally Posted by Revy

Just one thing, how do you get denominator in
$\frac{du}{dx} = \frac1{2\sqrt{1-x}\sqrt x} = \frac1{2\sqrt{x-x^2}}$ ?

$\sqrt{a} \cdot \sqrt{b} = \sqrt{a \cdot b}$

**Revy** · August 23rd 2010, 11:04 PM

Sorry, I should have asked more accurately.
I wanted to know why there is $2$ and not $1-2x$ from $d(x-x^2)=1-2x d(x-x^2)$
Do you differentiate $1-2x$ part? I don't understand the meaning of $\frac{du}{dx}$

Since we know formula $\arcsin{x} = \frac1{\sqrt{1-x^2}}$ how can you get $\arcsin{\sqrt{x}}$ from $\frac1{\sqrt{x-x^2}}$ ?

Shouldn't in this case equality be $\arcsin{\sqrt{x}} = \frac1{\sqrt{1-x}}$ ? If that's the case, then where does $\sqrt x$ disappear from $\frac1{2\sqrt{1-x}\sqrt x}$ and how $2$ gets from denominator to numerator?

Sorry, I must be a headache.

**tom@ballooncalculus** · August 24th 2010, 12:11 AM

Just in case a picture helps...

Differentiate arcsin root x (travelling downwards)...

... where (key in spoiler) ...

Spoiler:

Then, working backwards through the chain rule to integrate the whole of the original fraction...

... where (key in spoiler) ...

Spoiler:

Use the same diagram to check the integration by differentiating your result (i.e. by working down/clockwise instead of up/anti-clockwise through the same picture).
_________________________________________

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**Revy** · August 24th 2010, 03:45 AM

Thanks, it was hard, but seems I did understood it

**sa-ri-ga-ma** · August 24th 2010, 04:06 AM

Originally Posted by Revy

Thanks, it was hard, but seems I did understood it

General rule of integration is

$\int{f(x)f'(x)}dx = [f(x)]^2$ . Here f'(x) is the derivative of f(x)

In the given problem f(x) = arcsin(x) . Find f'(x)

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