Why we cant do it this way:

1/sq rt(1-v^2) = (1-v^2)^(-1/2)

Then integrate and it becomes (-1/3v)(1-v^2)^(3/2) ??

And how do we integrate sq rt[1+(2y)^2] with respect to y?

Can someone help me pls? Thanks!!!

- Thread starter makeAwish
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Why we cant do it this way:

1/sq rt(1-v^2) = (1-v^2)^(-1/2)

Then integrate and it becomes (-1/3v)(1-v^2)^(3/2) ??

And how do we integrate sq rt[1+(2y)^2] with respect to y?

Can someone help me pls? Thanks!!!

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I don't know what you have done exactly there, I think it is wrong though.1/sq rt(1-v^2) = (1-v^2)^(-1/2)

Then integrate and it becomes (-1/3v)(1-v^2)^(3/2) ??

The way I would integrate it:

[tex]\int\frac{1}{\sqrt{1-v^2}} dv[/tex]

[tex]v = sin(\vartheta)[/tex]

[tex]\vartheta = arcsin(v)[/tex]

[tex]\frac{dv}{d\vartheta}=cos(\vartheta)[/tex]

[tex]dv=cos(\vartheta)d\vartheta[/tex]

so...

[tex]\int\frac{1}{\sqrt{1-v^2}} dv = \int\frac{1}{cos(\vartheta)}}cos(\vartheta) dv = \vartheta + C = arcsin(v) + C[/tex]

how about:And how do we integrate sq rt[1+(2y)^2] with respect to y?

[tex]y = \frac{cos(\vartheta)}{2}[/tex]

edit: I meant sin(theta)/2

edit2: very sorry, I missread the question you should do what derek e said: let 2y = sinh(a).

Last edited:

hmm. what is sinh(a)?Yeah, use trig substitution.

Differentiate the integrated function in the second part. I don't think it checks out.

Let 2y = sinh(a).

sorry i have not learnt that yet..

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It is a hyperbolic function, more info is in the link.hmm. what is sinh(a)?

sorry i have not learnt that yet..

http://www.sosmath.com/trig/hyper/hyper01/hyper01.html

To stick with the functions you already know you could use the substitution I mentioned in my last post. If you use this substitution then later in your working you will need to use the identity

[tex]cos^2(\vartheta)=\frac{cos(2\vartheta)+1}{2}[/tex]

edit: I misread the question, you wont need that identity and my substitution doesn't help.

Last edited:

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I am not sure what exactly your method is.er.. i know this mtd is wrong, but i dont rly understand why. can someone explain to me pls? Thanks :)

1. you raised the power by 2 rather than 1. -1/2 + 1 = 1/2 not 3/2.

2. you divided by the derivative of the brackets, which would usually work, BUT the derivative has a variable in it so you cannot use this method. If you try to differentiate the answer that you got using your method, this 1/2v gets in the way because you have to differentiate that as well.

But as I said, I don't know what method you mean because you didn't raise the power by just one.

Sorry sorry, careless mistake, yah i actually meant the raising power that mtd.I am not sure what exactly your method is.

1. you raised the power by 2 rather than 1. -1/2 + 1 = 1/2 not 3/2.

2. you divided by the derivative of the brackets, which would usually work, BUT the derivative has a variable in it so you cannot use this method. If you try to differentiate the answer that you got using your method, this 1/2v gets in the way because you have to differentiate that as well.

But as I said, I don't know what method you mean because you didn't raise the power by just one.

Hmm.. So can i say like whenever we integrate sqrt of more than one terms, we always need to use trigo to solve? No matter is inverse sqrt or sqrt..

Okay. I tried and i gethow about:

[tex]y = \frac{cos(\vartheta)}{2}[/tex]

(say deter is x, for simplicity)

[tex]\int cos^{2}x dx[/tex]

= [tex]\int (1/2)(cos 2x +1) dx[/tex]

= (1/2)[(1/2)sin 2x +x]

But i can't find my intergation limits in terms of x.

Cos initially it is from 0 to 1,

at y=0, deter(which i call it x for simplicity) = 0

but at y= 1,

x= sin^(-1) 2

and sin^(-1) 2 is not valid? as in i typed in calculator, but error.

You were falling into the trap of forgetting the chain rule. You need to differentiate your final expression to see if it even makes sense. Once you do, you should realize that some of the expressions you got above are ridiculous.er.. i know this mtd is wrong, but i dont rly understand why. can someone explain to me pls? Thanks :)

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Under normal circumstances I would go through each step of the integration to show you how to do it, but I am worried that I would make a mistake and all of the maths professors that browse these forums would punish me with there huge brains.hmm.. but i still cant solve that integration..

integrate sq rt[1+(2y)^2] with respect to y.

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[tex]y = \frac{cosh(a)}{2}[/tex]

[tex]\frac{dy}{da}=\frac{sinh(a)}{2}[/tex]

[tex]\int\sqrt{1+4y^2}dy=\int\sqrt{1+4(\frac{cosh(a)}{2})^2}\frac{sinh(a)}{2}da[/tex]

[tex]= \int \frac{1}{2}sinh^2(a)da[/tex]

[tex]=\frac{1}{4}sinh(2a)-\frac{a}{2}+C[/tex]

[tex]a = arccosh(2y)[/tex]

(I can't be bothered to put it back in terms of y)

[itex]

\begin{equation*}\begin{split}

\int_0^1 \sqrt{1 + (2y)^2}dy &= \frac{1}{2}\int_\alpha^\beta cosh^2(a)da \\

&= \frac{1}{2}\int_\alpha^\beta \frac{1 + cosh(2a)}{2} da \\

&= \frac{1}{4}\left[a + \frac{1}{2}sinh(2a)\right]_\alpha^\beta

\end{split}\end{equation*}

[/itex]

Since it is a definite integral, this is a straightforward way of getting an answer.

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