Just verifying an integration by parts problem...

**Malaclypse** · February 5th 2011, 03:29 PM

Hey all, just checking my work on an integration by parts problem I've been working on. It's coming out to 0, which makes my spider sense tingle a little bit about the methodology I used to get there. It may be right, or it may be one of those situations where I've been staring at it for too long and it's really something obvious. I know I could have cancelled those terms earlier, but I just left them while I was working on it. What do you guys think? Is my method at least correct, and do you get the same? Thank you very much in advance.

$\begin{array}{l} \int {\frac{{\arcsin x}}{{\sqrt {1 + x} }}{\rm{d}}x} \\ \int {{{\sin }^{ - 1}}x\,{{(x + 1)}^{ - \frac{1}{2}}}} {\rm{d}}x\\ u = {\sin ^{ - 1}}x\\ {\rm{d}}u = \frac{1}{{\sqrt {1 - {x^2}} }}{\rm{d}}x\\ dv = {(x + 1)^{ - \frac{1}{2}}}\\ v = 2{(x + 1)^{\frac{1}{2}}}\\ \int {{{\sin }^{ - 1}}x\,{{(x + 1)}^{ - \frac{1}{2}}}} {\rm{d}}x = {\sin ^{ - 1}}x\left( {2\sqrt {x + 1} } \right) - 2\int {\frac{{\sqrt {x + 1} }}{{\sqrt {1 - {x^2}} }}{\rm{d}}x} \\ u = \sqrt {x + 1} \\ {\rm{d}}u = \frac{1}{2}{(x + 1)^{ - \frac{1}{2}}}{\rm{d}}x\\ dv = \frac{1}{{\sqrt {1 - {x^2}} }} \end{array}$
$\begin{array}{l} v = {\sin ^{ - 1}}x\\ \int {\frac{{{{\sin }^{ - 1}}x}}{{\sqrt {x + 1} }}\,} {\rm{d}}x = 2\left( {{{\sin }^{ - 1}}x} \right)\left( {\sqrt {x + 1} } \right) - 2\left[ {\left( {\sqrt {x + 1} } \right)\left( {{{\sin }^{ - 1}}x} \right) - \frac{1}{2}\int {\frac{{{{\sin }^{ - 1}}x}}{{\sqrt {x + 1} }}{\rm{d}}x} } \right]\\ \int {\frac{{{{\sin }^{ - 1}}x}}{{\sqrt {x + 1} }}\,} {\rm{d}}x = 2\left( {{{\sin }^{ - 1}}x} \right)\left( {\sqrt {x + 1} } \right) - 2\left( {\sqrt {x + 1} } \right)\left( {{{\sin }^{ - 1}}x} \right) - \int {\frac{{{{\sin }^{ - 1}}x}}{{\sqrt {x + 1} }}{\rm{d}}x} \\ 2\int {\frac{{{{\sin }^{ - 1}}x}}{{\sqrt {x + 1} }}\,} {\rm{d}}x = 2\left( {{{\sin }^{ - 1}}x} \right)\left( {\sqrt {x + 1} } \right) - 2\left( {\sqrt {x + 1} } \right)\left( {{{\sin }^{ - 1}}x} \right) \end{array}$
$\begin{array}{l} \int {\frac{{{{\sin }^{ - 1}}x}}{{\sqrt {x + 1} }}\,} {\rm{d}}x = \frac{{2\left( {{{\sin }^{ - 1}}x} \right)\left( {\sqrt {x + 1} } \right)}}{2} - \frac{{2\left( {\sqrt {x + 1} } \right)\left( {{{\sin }^{ - 1}}x} \right)}}{2}\\ \int {\frac{{{{\sin }^{ - 1}}x}}{{\sqrt {x + 1} }}\,} {\rm{d}}x = 0 \end{array}$

**tom@ballooncalculus** · February 5th 2011, 03:56 PM

OK to here...

$\displaystyle{-\ 2\int {\frac{{\sqrt {x + 1} }}{{\sqrt {1 - {x^2}} }}\ {\rm{d}}x}}$

... but then...

$\displaystyle{-\ 2\int {\frac{{\sqrt {x + 1} }}{{\sqrt {1 - {x^2}} }}\ {\rm{d}}x}}$

$\displaystyle{\ =\ -\ 2\int {\frac{{\sqrt {x + 1} }}{{\sqrt {(1 - x)(1 + x)}} }}\ {\rm{d}}x}}$

$\displaystyle{\ =\ -\ 2\int {\frac{{1}}{{\sqrt {(1 - x)}} }}\ {\rm{d}}x}}$

**Malaclypse** · February 5th 2011, 04:13 PM

Sneaky bugger...
$\int {{{\sin }^{ - 1}}x\,{{(x + 1)}^{ - \frac{1}{2}}}} {\rm{d}}x = {\sin ^{ - 1}}x\left( {2\sqrt {x + 1} } \right) - 2\int {\frac{{\sqrt {x + 1} }}{{\sqrt {(x + 1)} \sqrt {(x - 1)} }}{\rm{d}}x}$
$\int {{{\sin }^{ - 1}}x\,{{(x + 1)}^{ - \frac{1}{2}}}} {\rm{d}}x = {\sin ^{ - 1}}x\left( {2\sqrt {x + 1} } \right) - 2\int {{{(x - 1)}^{ - \frac{1}{2}}}{\rm{d}}x}$
$\begin{array}{l} u = x - 1\\ du = dx\\ \int {{u^{ - \frac{1}{2}}}{\rm{d}}u} \\ 2\sqrt u \\ 2\sqrt {x - 1} \end{array}$
$\int {{{\sin }^{ - 1}}x\,{{(x + 1)}^{ - \frac{1}{2}}}} {\rm{d}}x = {\sin ^{ - 1}}x\left( {2\sqrt {x + 1} } \right) - 4\sqrt {x - 1}$

That's looking a lot more correct. Is that what you see?

Also...I suppose that if you get a 0 by bringing the original integral back over (when you get one in a loop) and dividing the constant out it's a good hint that something further can be done?

Edit: Good thing this is a Math and not an English forum.

**tom@ballooncalculus** · February 5th 2011, 04:20 PM

Well it's a hint that your second half had been a futile undoing of the first half, if that's what you mean.

Watch your signs... why have you turned sqrt(1 - x) into sqrt(x - 1) ...?

**Malaclypse** · February 5th 2011, 04:28 PM

Because I've been doing caluclus for many, many hours and apparently have decided that I'd rather create my own mathematical rules than follow conventional ones.

Thanks.

I'll fix that too!

**tom@ballooncalculus** · February 5th 2011, 04:39 PM

Let's not be conventional! Let's do like...

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

Spoiler:

_________________________________________

Don't integrate - balloontegrate!

Balloon Calculus; standard integrals, derivatives and methods

Balloon Calculus Drawing with LaTeX and Asymptote!

**TheCoffeeMachine** · February 5th 2011, 04:43 PM

Originally Posted by Malaclypse

$\begin{array}{l} \int {\frac{{{{\sin }^{ - 1}}x}}{{\sqrt {x + 1} }}\,} {\rm{d}}x = 2\left( {{{\sin }^{ - 1}}x} \right)\left( {\sqrt {x + 1} } \right) - 2\left[ {\left( {\sqrt {x + 1} } \right)\left( {{{\sin }^{ - 1}}x} \right) - \frac{1}{2}\int {\frac{{{{\sin }^{ - 1}}x}}{{\sqrt {x + 1} }}{\rm{d}}x} } \right]\\ \int {\frac{{{{\sin }^{ - 1}}x}}{{\sqrt {x + 1} }}\,} {\rm{d}}x = 2\left( {{{\sin }^{ - 1}}x} \right)\left( {\sqrt {x + 1} } \right) - 2\left( {\sqrt {x + 1} } \right)\left( {{{\sin }^{ - 1}}x} \right) - \int {\frac{{{{\sin }^{ - 1}}x}}{{\sqrt {x + 1} }}{\rm{d}}x} \\ 2\int {\frac{{{{\sin }^{ - 1}}x}}{{\sqrt {x + 1} }}\,} {\rm{d}}x = 2\left( {{{\sin }^{ - 1}}x} \right)\left( {\sqrt {x + 1} } \right) - 2\left( {\sqrt {x + 1} } \right)\left( {{{\sin }^{ - 1}}x} \right) \end{array}$
$\begin{array}{l} \int {\frac{{{{\sin }^{ - 1}}x}}{{\sqrt {x + 1} }}\,} {\rm{d}}x = \frac{{2\left( {{{\sin }^{ - 1}}x} \right)\left( {\sqrt {x + 1} } \right)}}{2} - \frac{{2\left( {\sqrt {x + 1} } \right)\left( {{{\sin }^{ - 1}}x} \right)}}{2}\\ \int {\frac{{{{\sin }^{ - 1}}x}}{{\sqrt {x + 1} }}\,} {\rm{d}}x = 0 \end{array}$

It seems you have a sign error in the second step. You should have had:

$\displaystyle \int {\frac{{{{\sin }^{ - 1}}x}}{{\sqrt {x + 1} }}\,} {\rm{d}}x = 2\left( {{{\sin }^{ - 1}}x} \right)\left( {\sqrt {x + 1} } \right) - 2\left( {\sqrt {x + 1} } \right)\left( {{{\sin }^{ - 1}}x} \right) +\int {\frac{{{{\sin }^{ - 1}}x}}{{\sqrt {x + 1} }}{\rm{d}}x}.$

Which takes you back to exactly where you had started:

$\displaystyle \int {\frac{{{{\sin }^{ - 1}}x}}{{\sqrt {x + 1} }}\,} {\rm{d}}x = \int {\frac{{{{\sin }^{ - 1}}x}}{{\sqrt {x + 1} }}{\rm{d}}x}.$

In other words, what you did was basically:

$\displaystyle \begin{aligned}\int f'(x)g(x)\;{dx} & = f(x)g(x)\;{dx}-\int f(x)g'(x)\;{dx} \\& = f(x)g(x)-f(x)g(x)+\int f'(x)g(x)\;{dx} \\& = \int f'(x)g(x)\;{dx}.\end{aligned}$

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