Definite Integral Help

[Nazariy]

Hello everyone, I would greatly appreciate your help on this

Consider the following:

\(\displaystyle \displaystyle{\int_{\frac{-\pi}{8}}^{\frac{\pi}{8}} \frac{4\cos4t}{\sqrt{1+\sin4t}+\sqrt{1-\sin4t} + 2} dt} = 2\int_{0}^{\frac{\pi}{8}} \frac{4\cos4t}{(\cos2t+\sin2t)+(\cos2t-\sin2t) + 2} dt\)
​

I have a number of conceptual questions regarding this.

1) Is there a way to tell in a short amount of time, just thinking about it, that the fraction on the LHS is actually a symmetrical graph reflected in y axis? (because if there is such a way, then you can easily rewrite the integral on the left as two integrals of that fraction with the bounds \(\displaystyle 0\) to \(\displaystyle \displaystyle{\frac{\pi}{8}}\))

2) If there is no such way, then consider the fraction on the RHS. Can you tell from the fraction on the RHS that it is symmetrical (i.e. \(\displaystyle 4\cos4t/2\cos2t\))?

Suppose I have figured out that the fraction is a symmetrical graph in y axis.

3) How do I know that I should take a positive sqrt and not a negative when rewriting the sqrts. i.e. \(\displaystyle 1+\sin4t = \sin^2 2t + \cos^2 2t + 2\cos2t\sin2t\), therefore \(\displaystyle 1+\sin4t = (\cos2t+\sin2t)^2\), thus \(\displaystyle \sqrt{(\cos2t+\sin2t)^2}\), which can be either positive or negative \(\displaystyle (\cos2t + \sin2t)\). How do I know which one to take. The same question clearly applies to \(\displaystyle \sqrt{1-\sin4t}\).

4) Suppose I have figured out which sqrt sign to take. How do I know that \(\displaystyle \sqrt{1-\sin4t} = \cos2t - \sin2t\) and not \(\displaystyle \sin2t-cos2t\). Clearly \(\displaystyle (\cos2t-\sin2t)^2\) is the same as \(\displaystyle (\sin2t-\cos2t)^2\), however it makes a huge difference if you pick one or the other. This is because in the denominator on the RHS you either end up with \(\displaystyle 2\cos2t\) or \(\displaystyle 2\sin2t\), based on your choice, and two are clearly not the same.

If you could answer all/some of those... It would be an immense help.

Thank you

 

[Ishuda]

Hello everyone, I would greatly appreciate your help on this

Consider the following:

\(\displaystyle \displaystyle{\int_{\frac{-\pi}{8}}^{\frac{\pi}{8}} \frac{4\cos4t}{\sqrt{1+\sin4t}+\sqrt{1-\sin4t} + 2} dt} = 2\int_{0}^{\frac{\pi}{8}} \frac{4\cos4t}{(\cos2t+\sin2t)+(\cos2t-\sin2t) + 2} dt\)
​

I have a number of conceptual questions regarding this.

1) Is there a way to tell in a short amount of time, just thinking about it, that the fraction on the LHS is actually a symmetrical graph reflected in y axis? (because if there is such a way, then you can easily rewrite the integral on the left as two integrals of that fraction with the bounds \(\displaystyle 0\) to \(\displaystyle \displaystyle{\frac{\pi}{8}}\))

2) If there is no such way, then consider the fraction on the RHS. Can you tell from the fraction on the RHS that it is symmetrical (i.e. \(\displaystyle 4\cos4t/2\cos2t\))?

Suppose I have figured out that the fraction is a symmetrical graph in y axis.

3) How do I know that I should take a positive sqrt and not a negative when rewriting the sqrts. i.e. \(\displaystyle 1+\sin4t = \sin^2 2t + \cos^2 2t + 2\cos2t\sin2t\), therefore \(\displaystyle 1+\sin4t = (\cos2t+\sin2t)^2\), thus \(\displaystyle \sqrt{(\cos2t+\sin2t)^2}\), which can be either positive or negative \(\displaystyle (\cos2t + \sin2t)\). How do I know which one to take. The same question clearly applies to \(\displaystyle \sqrt{1-\sin4t}\).

4) Suppose I have figured out which sqrt sign to take. How do I know that \(\displaystyle \sqrt{1-\sin4t} = \cos2t - \sin2t\) and not \(\displaystyle \sin2t-cos2t\). Clearly \(\displaystyle (\cos2t-\sin2t)^2\) is the same as \(\displaystyle (\sin2t-\cos2t)^2\), however it makes a huge difference if you pick one or the other. This is because in the denominator on the RHS you either end up with \(\displaystyle 2\cos2t\) or \(\displaystyle 2\sin2t\), based on your choice, and two are clearly not the same.

If you could answer all/some of those... It would be an immense help.

Thank you

Depends on what you mean on short amount of time but it should be fairly obvious that f(-t)=f(t) where f is the function which represents the integrand on either the LHS or RHS. That is cos(-x)=cos(x) and sin(-x)=-sin(x).

As far as the square root goes, the convention is the square root in an equation represents the positive square root. That is \(\displaystyle \sqrt{x^2}\)=|x|. The choice of cos(2t) + sin(2t) and cos(2t) - sin(2t) follows that convention.

 

[Steven G]

Hello everyone, I would greatly appreciate your help on this

Consider the following:

\(\displaystyle \displaystyle{\int_{\frac{-\pi}{8}}^{\frac{\pi}{8}} \frac{4\cos4t}{\sqrt{1+\sin4t}+\sqrt{1-\sin4t} + 2} dt} = 2\int_{0}^{\frac{\pi}{8}} \frac{4\cos4t}{(\cos2t+\sin2t)+(\cos2t-\sin2t) + 2} dt\)
​

I have a number of conceptual questions regarding this.

1) Is there a way to tell in a short amount of time, just thinking about it, that the fraction on the LHS is actually a symmetrical graph reflected in y axis? (because if there is such a way, then you can easily rewrite the integral on the left as two integrals of that fraction with the bounds \(\displaystyle 0\) to \(\displaystyle \displaystyle{\frac{\pi}{8}}\))

2) If there is no such way, then consider the fraction on the RHS. Can you tell from the fraction on the RHS that it is symmetrical (i.e. \(\displaystyle 4\cos4t/2\cos2t\))?

Suppose I have figured out that the fraction is a symmetrical graph in y axis.

3) How do I know that I should take a positive sqrt and not a negative when rewriting the sqrts. i.e. \(\displaystyle 1+\sin4t = \sin^2 2t + \cos^2 2t + 2\cos2t\sin2t\), therefore \(\displaystyle 1+\sin4t = (\cos2t+\sin2t)^2\), thus \(\displaystyle \sqrt{(\cos2t+\sin2t)^2}\), which can be either positive or negative \(\displaystyle (\cos2t + \sin2t)\). How do I know which one to take. The same question clearly applies to \(\displaystyle \sqrt{1-\sin4t}\).

4) Suppose I have figured out which sqrt sign to take. How do I know that \(\displaystyle \sqrt{1-\sin4t} = \cos2t - \sin2t\) and not \(\displaystyle \sin2t-cos2t\). Clearly \(\displaystyle (\cos2t-\sin2t)^2\) is the same as \(\displaystyle (\sin2t-\cos2t)^2\), however it makes a huge difference if you pick one or the other. This is because in the denominator on the RHS you either end up with \(\displaystyle 2\cos2t\) or \(\displaystyle 2\sin2t\), based on your choice, and two are clearly not the same.

If you could answer all/some of those... It would be an immense help.

Thank you

Sometime |x| is x and other times it is -x. You need to use the limits of integration to determine when to use x or -x. Of course I am using the fact that sqrt(x^2)=|x|.
Consider the example
\displaystyle{\int_{\frac{-{3}}^{\frac{{8}} \|x| dx}

 

[Steven G]

Sometime |x| is x and other times it is -x. You need to use the limits of integration to determine when to use x or -x. Of course I am using the fact that sqrt(x^2)=|x|.
Consider the example int (-3,6) |x|dx = int (-3,0) (-x)dx + int (0,6) x dx

 

[Nazariy]

Depends on what you mean on short amount of time but it should be fairly obvious that f(-t)=f(t) where f is the function which represents the integrand on either the LHS or RHS. That is cos(-x)=cos(x) and sin(-x)=-sin(x).

As far as the square root goes, the convention is the square root in an equation represents the positive square root. That is \(\displaystyle \sqrt{x^2}\)=|x|. The choice of cos(2t) + sin(2t) and cos(2t) - sin(2t) follows that convention.

The answer to this question mentioned inspecting what happens to the function in the range of integration. And the function is symmetric around y axis and is positive, and that is the reason why they pick a positive sqrt, I suppose it makes sense (they also mention that cos2t > sin2t for values of t in the range of integration, is that why I would pick cos2t-sin2t over sin2t-cos2t?). But what if a combination of different sign sqrts yields the positive result too... For example if I take a negative square root of \(\displaystyle \sqrt{1-\sin4t}\) and positive of \(\displaystyle \sqrt{1+\sin4t}\) and still get positive overall value for the fraction in the whole of the integration range (clearly that must not be the case in this instance, and this would take way too long to figure out when in an exam; but I suppose that would have to be solution too?).

How come the convention is that, if you can have \(\displaystyle -(x)*-(x) = x^2\). And then clearly \(\displaystyle \sqrt{x^2}=\pm x\). It is just that, a convention, I still have to make a judgement on which sign to use, or do I just blindly go for positive all the time?

 

[Nazariy]

Sometime |x| is x and other times it is -x. You need to use the limits of integration to determine when to use x or -x. Of course I am using the fact that sqrt(x^2)=|x|.
Consider the example
\displaystyle{\int_{\frac{-{3}}^{\frac{{8}} \|x| dx}

Your code is definitely a bit off..

So how would I judge what sign to take, taking into account what happens to the function in that integration range?

Can you please re post your example. I would be very thankful!

 

[Ishuda]

...

How come the convention is that, if you can have \(\displaystyle -(x)*-(x) = x^2\). And then clearly \(\displaystyle \sqrt{x^2}=\pm x\). It is just that, a convention, I still have to make a judgement on which sign to use, or do I just blindly go for positive all the time?

CUZ that's the convention It isn't blindly, it is the convention that
\(\displaystyle \sqrt{x^2}\, =\, |x|\)
AND, as mentioned above by Jomo, it is important to note the limits of integration. That is
\(\displaystyle \sqrt{1\, -\, sin(4t)}\, =\, cos(2t)\, -\, sin(2t);\, t\, \epsilon\, [-\frac{\pi}{8},\, \frac{\pi}{8}]\)
However,
\(\displaystyle \sqrt{1\, -\, sin(4t)}\, =\, sin(2t)\, -\, cos(2t);\, t\, \epsilon\, [\frac{\pi}{4},\, \frac{\pi}{2}]\)

 

[Nazariy]

CUZ that's the convention It isn't blindly, it is the convention that
\(\displaystyle \sqrt{x^2}\, =\, |x|\)
AND, as mentioned above by Jomo, it is important to note the limits of integration. That is
\(\displaystyle \sqrt{1\, -\, sin(4t)}\, =\, cos(2t)\, -\, sin(2t);\, t\, \epsilon\, [-\frac{\pi}{8},\, \frac{\pi}{8}]\)
However,
\(\displaystyle \sqrt{1\, -\, sin(4t)}\, =\, sin(2t)\, -\, cos(2t);\, t\, \epsilon\, [\frac{\pi}{4},\, \frac{\pi}{2}]\)

Boom Since my beloved convention is \(\displaystyle \sqrt{x^2}\, =\, |x|\), it means that I have to go for \(\displaystyle \cos(2t)-\sin(2t)\) because if you plot both, you will see that \(\displaystyle \cos(2t) > \sin(2t)\) for every value of t in that range. Now I understand that and I will be on a lookout for something like this in the future problems that I will solve. However, I still don't like following the convention, something like this should be explicitly mentioned in the question so as to avoid any ambiguity for the solver... Negative values should not be discriminated against... I mean, that integral can be solved in 4 ways due to combinations of square root, if we are to ignore the convention

It would also be quite useful to see some other examples of this convention been used in practice, do you by any chance know of a resource that I can have a read through, if there were some exercises, it wouldn't be too shabby as well.

You have helped me a lot, so if you deem this question has fulfilled its purpose, please feel free to ignore my reply to you

Thank you Ishuda!

 

[Ishuda]

Boom Since my beloved convention is \(\displaystyle \sqrt{x^2}\, =\, |x|\), it means that I have to go for \(\displaystyle \cos(2t)-\sin(2t)\) because if you plot both, you will see that \(\displaystyle \cos(2t) > \sin(2t)\) for every value of t in that range. Now I understand that and I will be on a lookout for something like this in the future problems that I will solve. However, I still don't like following the convention, something like this should be explicitly mentioned in the question so as to avoid any ambiguity for the solver... Negative values should not be discriminated against... I mean, that integral can be solved in 4 ways due to combinations of square root, if we are to ignore the convention

It would also be quite useful to see some other examples of this convention been used in practice, do you by any chance know of a resource that I can have a read through, if there were some exercises, it wouldn't be too shabby as well.

You have helped me a lot, so if you deem this question has fulfilled its purpose, please feel free to ignore my reply to you

Thank you Ishuda!

Lets go back to algebra days and the good old quadratic equation:
a x² + b x + c = 0, a \(\displaystyle \ne\) 0.
The solutions were given by
x = \(\displaystyle \frac{-b\, \pm\, \sqrt{b^2\, -\, 4\, a\, c}}{2\, a}\)
That \(\displaystyle \pm\) is there because, by convention, the square root [whether \(\displaystyle \sqrt{x^2}\) or (x²)^1/2] means the positive square root, i.e. |x|.