Integration (1 Viewer)

Drongoski · May 2, 2009

jm01 said:
Thanks gurmies...

I have a little problem again here...

The question was:

$\int (9-x^2)^\frac{-1}{2} dx$

The answer in the back was:

$\frac{1}{2}( x \sqrt{9-x^2} + 9sin^{-1}\frac{x}{3})$

I put it through Wolfram Mathematica and got:

$\frac{1}{2} x \sqrt{9+x^2}+\frac{9}{2} \text{ArcSinh}\left[\frac{x}{3}\right]$

I thought the answer would just be:
$sin^{-1}\frac{x}{3}$

Which is the correct answer, and what is ArcSinh??

Thanks

Looks like yr answer sin^-1(x/3) + C is correct. When u differentiate it it gives back the integrand.

lolokay · May 2, 2009

arcsinh is sinh^-1 which is inverse hyperbolic sine. you put 9+x² instead of minus

are you sure the power is written as -1/2 rather than just 1/2?

http://integrals.wolfram.com/index.jsp?expr=(9-x^2)^(1/2)&random=false gives the books answer

Drongoski · May 2, 2009

Hyperbolic functions of x

$\100dpi sinh\ x\ \ =\ \ \frac{e^{x}-e^{-x}}{2}\ \ \ \cdots \ \ \ sin\ x\ \ =\ \ \frac{e^{ix}-e^{-ix}}{2i}\\ \\ \ \ cosh\ x\ \ =\ \ \frac{e^{x}+e^{-x}}{2}\ \ \ \cdots \ \ \ cos\ x\ \ =\ \ \frac{e^{ix}+e^{-ix}}{2}\\ \\ \ \ tanh\ x\ \ =\ \ \frac{sinh\ x}{cosh\ x}\\ \\ sinh\ 2x\ \ =\ \ 2\ sinh\ x\ cosh\ x\\ \\ cosh^{2}\ x\ \ -\ \ sinh^{2}\ x\ \ =\ \ 1\\ \\ etc\ \ etc\ \$

sinh^-1( x) is just the hyperbolic analogue of circular version sin^-1 (x)

cutemouse · May 2, 2009

Yes I typed the question wrong... -_-

The correct question is:

$\int (9-x^2)^\frac{1}{2}$

Can anybody do it please?

Thanks =D

azureus88 · May 2, 2009

[maths]$Let$\ x=3\sin\phi \\dx=3\cos\phi\, d\phi[/maths]

[maths]\int \sqrt{9-x^2}dx\\=\int (3\cos\phi )(3\cos\phi )d\phi \\=9\int \cos^2\phi d\phi \\=\frac{9}{2}\int (1+\cos2\phi) d\phi \\=\frac{9}{2}(\phi +\frac{1}{2}\sin2\phi )+c\\=\frac{9}{2}\sin^{-1}\frac{x}{3}+\frac{9}{2}(\frac{x}{3})(\frac{\sqrt{9-x^2}}{3})+c\\=\frac{9}{2}\sin^{-1}\frac{x}{3}+\frac{x\sqrt{9-x^2}}{2}+c[/maths]

Alternatively, you could do it by parts if you don't lik substituting stuff back in.

cutemouse · May 3, 2009

Hi, I've yet again have one more question...

$\int(9+x^2)^\frac{1}{2}dx$

Thanks

Trebla · May 3, 2009

Assuming you've typed the question correctly, you should be able to do it yourself. Just expand it and integrate like usual...

cutemouse · May 3, 2009

*doh* Thanks Trebla... I'm really off the pace today

Edited with the correct question

lolokay · May 3, 2009

do it by parts

$\int \sqrt{9+x^2}dx \\ =x\sqrt{9+x^2} - \int \frac{x^2}{\sqrt{9+x^2}}dx\\ =x\sqrt{9+x^2} - \int \frac{x^2+9}{\sqrt{9+x^2}}dx + 9\int\frac{dx}{\sqrt{9+x^2}}dx\\ = x\sqrt{9+x^2} - \int \sqrt{9+x^2}dx + 9ln(x+\sqrt{9+x^2})\\ 2\int \sqrt{9+x^2}dx = x\sqrt{9+x^2} + 9ln(x+\sqrt{9+x^2})\\ \int \sqrt{9+x^2}dx = \frac{1}{2}(x\sqrt{9+x^2} + 9ln(x+\sqrt{9+x^2}))+C$

cutemouse · May 3, 2009

Hi Thanks for that...

I have another one from hell

Could someone please help me? Thanks

$\int ln(x+\sqrt{x^2-1})dx$

And yes this is the correct one, I've double checked!

lolokay · May 3, 2009

again, just do it by parts

d/dx ln(x+sqrt[x^2-1]) = 1/sqrt[x^2-1]
then just a basic application of ibp

cutemouse · May 3, 2009

Thanks for that... I have one more question

As you can tell it's the weekend, and I'm working through exercises...

$\int_1^2 \frac{(x+1)dx}{\sqrt{-2+3x-x^2}}$

Thanks

study-freak · May 3, 2009

Trebla · May 3, 2009

jm01 said:
Thanks for that... I have one more question As you can tell it's the weekend, and I'm working through exercises...

$\int_1^2 \frac{(x+1)dx}{\sqrt{-2+3x-x^2}}$

Thanks

Hint:

$\int_1^2 \frac{(x+1)}{\sqrt{-2+3x-x^2}}\,dx=\int_1^2 -\frac{1}{2}\frac{-2x+3-5}{\sqrt{-2+3x-x^2}\,dx}\\=\int_1^2 -\frac{1}{2}\frac{-2x+3}{\sqrt{-2+3x-x^2}\,dx}+\int_1^2 -\frac{1}{2}\frac{-5}{\sqrt{-2+3x-x^2}\,dx}$

gurmies · May 3, 2009

Yeah dude, practise being able to transform stuff like Trebla showed you, it's really useful and logical.

Dumbledore · May 3, 2009

Trebla said:
Hint:

$\int_1^2 \frac{(x+1)}{\sqrt{-2+3x-x^2}}\,dx=\int_1^2 -\frac{1}{2}\frac{-2x+3-5}{\sqrt{-2+3x-x^2}\,dx}\\=\int_1^2 -\frac{1}{2}\frac{-2x+3}{\sqrt{-2+3x-x^2}\,dx}-\int_1^2 -\frac{1}{2}\frac{-5}{\sqrt{-2+3x-x^2}\,dx}$

i don't think many people will get it if they have never seen this method b4

gurmies · May 3, 2009

True, but it's quite beneficial to learn to use and recognise.

Trebla said:
Hint:

$\int_1^2 \frac{(x+1)}{\sqrt{-2+3x-x^2}}\,dx=\int_1^2 -\frac{1}{2}\frac{-2x+3-5}{\sqrt{-2+3x-x^2}\,dx}\\=\int_1^2 -\frac{1}{2}\frac{-2x+3}{\sqrt{-2+3x-x^2}\,dx}-\int_1^2 -\frac{1}{2}\frac{-5}{\sqrt{-2+3x-x^2}\,dx}$

Think there should be a + sign between the two integrals...

cutemouse · May 3, 2009

Trebla said:
Hint:

$\int_1^2 \frac{(x+1)}{\sqrt{-2+3x-x^2}}\,dx=\int_1^2 -\frac{1}{2}\frac{-2x+3-5}{\sqrt{-2+3x-x^2}\,dx}\\=\int_1^2 -\frac{1}{2}\frac{-2x+3}{\sqrt{-2+3x-x^2}\,dx}-\int_1^2 -\frac{1}{2}\frac{-5}{\sqrt{-2+3x-x^2}\,dx}$

Hi, I tried it that way, but I get ln0 for some reason when I evaluate it, which does not exist. How do I work around this?

Thanks

Trebla · May 3, 2009

Ok, i just edited a minor error as mentioned by gurmies.

$\int_1^2 \frac{(x+1)}{\sqrt{-2+3x-x^2}}\,dx=\int_1^2 -\frac{1}{2}\frac{-2x+3-5}{\sqrt{-2+3x-x^2}\,dx}\\=\int_1^2 -\frac{1}{2}\frac{-2x+3}{\sqrt{-2+3x-x^2}\,dx}+\int_1^2 -\frac{1}{2}\frac{-5}{\sqrt{-2+3x-x^2}\,dx}$

The first integral does NOT give a logarithm. Apply the reverse chain rule or use the substitution u = - 2 + 3x - x².

For the second integral, complete the square of the expression inside the square root and you end up with a standard integral.

cutemouse · May 3, 2009

Thanks Trebla... my bad...

Sorry guys but another question... This one's annoying as when I'm trying to fiddle around with trig integrals, I have to deal with tan(pi/2) not being defined and cot0 not being defined

$\int_0^\frac{\pi}{2} \frac{dx}{cos^2x+2sin^2x}$

Thanks

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