Integral Problem?

May 2010
16
0
What have I done wrong here?

(The integral is that in the red square here & the denominator is supposed to be under a square root)



Which way is the correct way, and why???
 
Apr 2010
384
153
Canada
What have I done wrong here?

(The integral is that in the red square here & the denominator is supposed to be under a square root)



Which way is the correct way, and why???
\(\displaystyle \int \frac{dx}{\sqrt{ 4^2 - x^2 } } \)

Let \(\displaystyle x = 4sinp \) and \(\displaystyle dx = 4cosp \)

\(\displaystyle \int \frac{4cosp}{ \sqrt{ 16(1-sin^2p) }} \)

\(\displaystyle \int \frac{cosp}{ cosp } dp \)

\(\displaystyle p \)

\(\displaystyle p = sin^{-1} \frac{x}{4} \)
 
May 2010
16
0
\(\displaystyle \int \frac{dx}{\sqrt{ 4^2 - x^2 } } \)

Let \(\displaystyle x = 4sinp \) and \(\displaystyle dx = 4cosp \)

\(\displaystyle \int \frac{4cosp}{ \sqrt{ 16(1-sin^2p) }} \)

\(\displaystyle \int \frac{cosp}{ cosp } dp \)

\(\displaystyle p \)

\(\displaystyle p = sin^{-1} \frac{x}{4} \)
Hey, thanks for reading that.

I have no idea why you chose sinp, why not cosp?

If you look at the triangle's I have above, in a lecture I seen (at 01:20 to 01:45) the teacher said that's where this method comes from & that is the thing to use, but which way is right?

The teacher said choosing either way works, i.e. labelling the legs of the triangle doesn't matter, but my example here clearly shows otherwise!
 
Apr 2010
384
153
Canada
Hey, thanks for reading that.

I have no idea why you chose sinp, why not cosp?

If you look at the triangle's I have above, in a lecture I seen (at 01:20 to 01:45) the teacher said that's where this method comes from & that is the thing to use, but which way is right?

The teacher said choosing either way works, i.e. labelling the legs of the triangle doesn't matter, but my example here clearly shows otherwise!
You are taking advantage of the identity \(\displaystyle sin^2 x + cos^2 x = 1 \) so you can use either cos or sin here.

We use this type of substitution when we have something like,

1: \(\displaystyle \sqrt{ a^2 - x^2} \)

2: \(\displaystyle \frac{ 1}{ \sqrt{ a^2 - x^2 }} \)

It's just what we do to make it easiar. Find more at: Trigonometric substitutions
 
May 2010
16
0
Thanks, I get that part it's just the one little thing I don't understand.

If you look at the picture I made of the two triangles, you see that I've labelled them both differently.

When I did all the work I got the same answer until I evaluated the angle, I just need to understand why you do it one way & not the other way.
 
Apr 2010
384
153
Canada
Thanks, I get that part it's just the one little thing I don't understand.

If you look at the picture I made of the two triangles, you see that I've labelled them both differently.

When I did all the work I got the same answer until I evaluated the angle, I just need to understand why you do it one way & not the other way.
To be honest I havent looked at your diagram to hard. It's killing me to look at it, but the answer should be the same. Let's compute using cos subsitution

\(\displaystyle \int \frac{dx}{ \sqrt{ 4^2 - x^2}} \)

\(\displaystyle x = 4 cos p \)

\(\displaystyle = - \int \frac{ sinp }{sinp} dp \)

\(\displaystyle = -p + c \)

\(\displaystyle = - cos^{-1} \frac{x}{4} + c \)

Note the negative. This should make your equations equal.
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
Thanks, I get that part it's just the one little thing I don't understand.

If you look at the picture I made of the two triangles, you see that I've labelled them both differently.

When I did all the work I got the same answer until I evaluated the angle, I just need to understand why you do it one way & not the other way.
You don't have to do it one way & not the other way. Both the sine and cosine substitutions work and give the same answer- except that you have dropped the "-" on the left side. You should get \(\displaystyle -cos^{-1}(\frac{\pi}{4})+ C\) and \(\displaystyle sin^{-1}(\frac{\pi}{4})+ D\). Note also that I have "C" and "D" where you have two "C"s. You have no reason to thing that the two constants of integration will be the same.

You got different answers when you "evaluated the angles" because you are looking at different angles. If \(\displaystyle A= cos\theta)\) and \(\displaystyle A= sin(\phi)\) then \(\displaystyle \theta+ \phi= \frac{pi}{2}\). That is, \(\displaystyle cos^{-1}(A)+ sin^{-1}(A)= \frac{\pi}{2}\) so that \(\displaystyle sin^{-1}(A)= -cos^{-1}(A)+ \frac{\pi}{2}\). Your two answers are exactly the same except for the constant of integration.
 
Last edited:
May 2010
16
0
Ah yes, that's great! That makes a lot more sense.

I just forgot to write the minus out in front of the cosine at the end even though I included it in the integration (Doh).

Thanks AllanCuz, I get it now.

I'm just wondering, HallsofIvy, if you understand what I'm doing drawing this right triangle & then integrating the whole equation from it?

If you do I wonder is this the easy way to integrate functions that will end up as an inverse sine or an inverse arctangent without memorizing all those ridiculous formulas?

I know that these formula's are recognisable from a square root of two perfect squares in the denominator but they are all so similar that just blatant memorising is bound to fail.

Could you let me know if this is the best way to deal with these functions & if there are slip ups or failings?