# Just verifying an integration by parts problem...

• Feb 5th 2011, 03:29 PM
Malaclypse
Just verifying an integration by parts problem...
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.

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

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

$$\begin{array}{l} \int {\frac{{{{\sin }^{ - 1}}x}}{{\sqrt {x + 1} }}\,} {\rm{d}}x = \frac{{2\left( {{{\sin }^{ - 1}}x} \right)\left( {\sqrt {x + 1} } \right)}}{2} - \frac{{2\left( {\sqrt {x + 1} } \right)\left( {{{\sin }^{ - 1}}x} \right)}}{2}\\ \int {\frac{{{{\sin }^{ - 1}}x}}{{\sqrt {x + 1} }}\,} {\rm{d}}x = 0 \end{array}$$
• Feb 5th 2011, 03:56 PM
tom@ballooncalculus
OK to here...

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

... but then...

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

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

$\displaystyle{\ =\ -\ 2\int {\frac{{1}}{{\sqrt {(1 - x)}} }}\ {\rm{d}}x}}$
• Feb 5th 2011, 04:13 PM
Malaclypse
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.
• Feb 5th 2011, 04:20 PM
tom@ballooncalculus
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) ...?
• Feb 5th 2011, 04:28 PM
Malaclypse
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!
• Feb 5th 2011, 04:39 PM
tom@ballooncalculus
(Rofl)

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

http://www.ballooncalculus.org/draw/parts/eleven.png

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

Spoiler:
http://www.ballooncalculus.org/asy/prod.png

... is the product rule. Straight continuous lines differentiate downwards (integrate up) with respect to x. And,

http://www.ballooncalculus.org/asy/maps/parts.png

... is lazy integration by parts, doing without u and v.

_________________________________________

Don't integrate - balloontegrate!

Balloon Calculus; standard integrals, derivatives and methods

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• Feb 5th 2011, 04:43 PM
TheCoffeeMachine
Quote:

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}