How can we use trigonometric substitution to integrate square root expressions?

[makeAwish]

To integrate 1/sq rt(1-v^2) with respect to v, we need to use trigo substitution to integrate rite?

Why we can't 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!

 

[derek e]

Yeah, use trig substitution.

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

Let 2y = sinh(a).

 

[Georgepowell]

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

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

I don't know what you have done exactly there, I think it is wrong though.

The way I would integrate it:

$$\int\frac{1}{\sqrt{1-v^2}} dv$$

$$v = sin(\vartheta)$$

$$\vartheta = arcsin(v)$$

$$\frac{dv}{d\vartheta}=cos(\vartheta)$$

$$dv=cos(\vartheta)d\vartheta$$
so...

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

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

how about:

$$y = \frac{cos(\vartheta)}{2}$$

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

 

[makeAwish]

Is it when integrating sq rt with more than one terms inside, we always need to use trigo to solve?

 

[makeAwish]

what i meant was:

$$\int 1/\sqrt{1-v^{2}} dv$$

= $$\int (1-v^{2})^{-1/2} dv$$

= $$\ (2/3)*(1/2v)(1-v^{2})^{3/2} + C$$

 

[makeAwish]

er.. i know this mtd is wrong, but i don't rly understand why. can someone explain to me pls? Thanks :)

 

[makeAwish]

Yeah, use trig substitution.

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

Let 2y = sinh(a).

hmm. what is sinh(a)?
sorry i have not learned that yet..

 

[Georgepowell]

hmm. what is sinh(a)?
sorry i have not learned that yet..

It is a hyperbolic function, more info is in the link.
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

$$cos^2(\vartheta)=\frac{cos(2\vartheta)+1}{2}$$

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

 

[Georgepowell]

er.. i know this mtd is wrong, but i don't rly understand why. can someone explain to me pls? Thanks :)

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.

 

[makeAwish]

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.

Sorry sorry, careless mistake, yah i actually meant the raising power that mtd.

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..

 

[makeAwish]

how about:

$$y = \frac{cos(\vartheta)}{2}$$

Okay. I tried and i get
(say deter is x, for simplicity)

$$\int cos^{2}x dx$$

= $$\int (1/2)(cos 2x +1) dx$$

= (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.

 

[makeAwish]

Actually for the integration of sq rt[1+(2y)^2] with respect to y, the integral limits are from 0 to 1. Sorry i realized i din mention the limits previously.

 

[snipez90]

er.. i know this mtd is wrong, but i don't rly understand why. can someone explain to me pls? Thanks :)

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.

 

[makeAwish]

hmm.. but i still can't solve that integration..

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

 

[Cyosis]

Use the relation $$\cosh^2 x - \sinh^2 x=1$$ and the substitution $$2y=\sinh x$$.

 

[Georgepowell]

hmm.. but i still can't solve that integration..

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

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.

 

[Georgepowell]

OK, I tried but I am pretty sure it is wrong. Is it wrong? If it is then why?:

$$y = \frac{cosh(a)}{2}$$

$$\frac{dy}{da}=\frac{sinh(a)}{2}$$

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

$$= \int \frac{1}{2}sinh^2(a)da$$

$$=\frac{1}{4}sinh(2a)-\frac{a}{2}+C$$

$$a = arccosh(2y)$$

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

 

[derek e]

Let 2y = sinh(a). Let $\alpha=sinh^{-1}(0)$ and $\beta=sinh^{-1}(2)$. Then
$\begin{equation*}\begin{split} \int_0^1 \sqrt{1 + (2y)^2}dy &= \frac{1}{2}\int_\alpha^\beta cosh^2(a)da \\ &= \frac{1}{2}\int_\alpha^\beta \frac{1 + cosh(2a)}{2} da \\ &= \frac{1}{4}\left[a + \frac{1}{2}sinh(2a)\right]_\alpha^\beta \end{split}\end{equation*}$
Since it is a definite integral, this is a straightforward way of getting an answer.