Do we need to memorise this identity? (3 Viewers)

SammyT123 · May 23, 2016

Solving the following integral:

$\\\int{\frac{dx}{x\sqrt{a^2+x^2}}}\\$ I now have $ I=\frac{1}{a}ln(tan\frac{\theta}{2}) $ which is correct $\\ \\ $To bring back the 'x' do I need the tan(x/2) = sqrt()/sqrt() result , I think its Cosine in the brackets?$\\ $Any other way to do it?$$

I subbed x=tan(theta)
t=atan(theta/2)

InteGrand · May 23, 2016

SammyT123 said:
Solving the following integral:

$\\\int{\frac{dx}{x\sqrt{a^2+x^2}}}\\$ I now have $ I=\frac{1}{a}ln(tan\frac{\theta}{2})\\ $To simplify do I need the tan(x/2) = sqrt()/sqrt() result -->Cosine?$\\ $Any other way to do it?$$

Which identity? You will need to leave the final answer in terms of x.

For that integral, you could also use a trig. substitution of x = a*tan(theta).

You end up with something involving integral of cosec(theta), which is:

-ln(cosec(theta) + cot(theta)).

Then you could convert this back to in terms of x, using trig. identities or a right-angled triangle to assist you. This avoids use of tan(theta/2).

leehuan · May 23, 2016

Rule of thumb with sqrt(a²+x²) is to sub x=a.tan(theta)

No memorising here aside from that^

SammyT123 · May 23, 2016

Sorry, I did substitute x=atan(theta)

Identity is give here
https://www.quora.com/How-could-one...ot-of-1-cosx-1+cosx-absolute-value-of-tan-x-2

SammyT123 · May 23, 2016

So my qs is

put tan(theta/2) in terms of x where x=tan(theta)

InteGrand · May 23, 2016

SammyT123 said:
So my qs is

put tan(theta/2) in terms of x where x=tan(theta)

$\noindent To get that identity from the Quora link, note that $\sin^2 \frac{a}{2}=\frac{1}{2}-\frac{1}{2}\cos a$ and $\cos^2 \frac{a}{2} = \frac{1}{2} + \frac{1}{2}\cos a$. Division gives $\tan^2 \frac{a}{2} = \frac{1-\cos a}{1+\cos a}$, provided all denominators are non-zero. Now, taking square roots yields $\left | \tan \frac{a}{2}\right |= \sqrt{ \frac{1-\cos a}{1+\cos a}}$.$

This identity isn't expected to be memorised for the HSC, but you should be able to derive it if needed.

SammyT123 · May 23, 2016

InteGrand said:
$\noindent To get that identity from the Quora link, note that $\sin^2 \frac{a}{2}=\frac{1}{2}-\frac{1}{2}\cos a$ and $\cos^2 \frac{a}{2} = \frac{1}{2} + \frac{1}{2}\cos a$. Division gives $\tan^2 \frac{a}{2} = \frac{1-\cos a}{1+\cos a}$, provided all denominators are non-zero. Now, taking square roots yields $\left | \tan \frac{a}{2}\right |= \sqrt{ \frac{1-\cos a}{1+\cos a}}$.$

This identity isn't expected to be memorised for the HSC, but you should be able to derive it if needed.

Could derive it, but imagine me sitting in the 4U trials thinking

x=tan(theta)
t=tan(theta/2)

"Oh boy better derive that weird squareroot cosine formula!!"

haha so is there any clue in the question or solution that allows me to conclude that formula is needed

Assuming I do derive this in the exam, the integral is still quite tedious

InteGrand · May 23, 2016

SammyT123 said:
Could derive it, but imagine me sitting in the 4U trials thinking

x=tan(theta)
t=tan(theta/2)

"Oh boy better derive that weird squareroot cosine formula!!"

haha so is there any clue in the question or solution that allows me to conclude that formula is needed

Well we didn't need that formula to do that integral in the original post. We could avoid half-angles as I showed. But if you did introduce half-angles, the way to get out of them is to use the half-angle formulas in some way, because these formulas tell us how to go from sin(a/2) and cos(a/2) (and hence tan(a/2)) to trig functions of a. (That's how I derived that formula above.)

SammyT123 · May 23, 2016

InteGrand said:
Which identity? You will need to leave the fnal answer in terms of x.

For that integral, you could also use a trig. substitution of x = a*tan(theta).

You end up with something involving integral of cosec(theta), which is:

-ln(cosec(theta) + cot(theta)).

Then you could convert this back to in terms of x, using trig. identities or a right-angled triangle to assist you. This avoids use of tan(theta/2).

Seem to be getting $\frac{1}{a}\int{\frac{1}{sin\theta}}$
Any way I can avoid half angle results?

See attached images in my previous post
(sorry I understand my explanation of my issue is not too great)

InteGrand · May 23, 2016

SammyT123 said:
Seem to be getting $\frac{1}{a}\int{\frac{1}{sin\theta}}$

$\noindent Yeah, that's $\csc \theta$. The integral of this is well known to be $-\ln \left|\csc \theta + \cot \theta \right|$, or another way to give it is $\ln \left | \csc \theta - \cot \theta \right |$. This can be shown by differentiating these.$

SammyT123 · May 23, 2016

InteGrand said:
$\noindent Yeah, that's $\csc \theta$. The integral of this is well known to be $-\ln \left|\csc \theta + \cot \theta \right|$, or another way to give it is $\ln \left | \csc \theta - \cot \theta \right |$. This can be shown by differentiating these.$

Ohh ok - Will make sure to remember that

Teacher told me its not a standard integral, so its best to use t-formula
If you dont mind, can you show me how I can finish off the solution from $\\ I = \frac{1}{a}ln(tan\frac{\theta}{2}) + C$
Given t=tan(theta/2)

(just curious thats all)

Paradoxica · May 23, 2016

SammyT123 said:
Solving the following integral:

$\\\int{\frac{dx}{x\sqrt{a^2+x^2}}}\\$ I now have $ I=\frac{1}{a}ln(tan\frac{\theta}{2}) $ which is correct $\\ \\ $To bring back the 'x' do I need the tan(x/2) = sqrt()/sqrt() result , I think its Cosine in the brackets?$\\ $Any other way to do it?$$

I subbed x=tan(theta)
t=atan(theta/2)

Pretty sure you can just back-differentiate to obtain the answer.

leehuan · May 23, 2016

T-formulae does allow the integral of cosec to come out nice and tidy, but there is no need for it at all

InteGrand · May 23, 2016

SammyT123 said:
Ohh ok - Will make sure to remember that
Teacher told me its not a standard integral, so its best to use t-formula
If you dont mind, can you show me how I can finish off the solution from $\frac{1}{a}ln(tan\frac{\theta}{2})$
Given t=tan(theta/2)

(just curious thats all)

$\noindent Yeah the integrals of sec and csc are not standard (for HSC purposes). But it's worth remembering them and how to quickly derive them (so that you can avoid half-angles). Here's the classic tricks:$

$\noindent We have$

$\begin{align*}\int \sec x \text{ d}x &= \int \frac{\sec x \left(\sec x + \tan x \right)}{\sec x + \tan x}\text{ d}x \\ &= \int \frac{\sec^2 x + \sec x \tan x}{\sec x + \tan x}\text{ d}x \\ &= \int \frac{\left(\sec x + \tan x\right)^{\prime}}{\sec x + \tan x}\text{ d}x \\ &= \ln \left |\sec x + \tan x \right |+c. \end{align*}$

$\noindent The trick for getting $\int \csc x \text{ d}x = -\ln \left|\csc x + \cot x \right | + c = \ln \left |\csc x - \cot x \right | + c$ is similar. Multiply top and bottom of the integrand by $\csc x + \cot x$ and observe that the numerator becomes the negative derivative of the denominator (or multiply top and bottom by $\csc x - \cot x$ and observe the numerator becomes the derivative of the denominator).$

SammyT123 · May 23, 2016

InteGrand said:
$\noindent Yeah the integrals of sec and csc are not standard (for HSC purposes). But it's worth remembering them and how to quickly derive them (so that you can avoid half-angles). Here's the classic tricks:$

$\noindent We have$

$\begin{align*}\int \sec x \text{ d}x &= \int \frac{\sec x \left(\sec x + \tan x \right)}{\sec x + \tan x}\text{ d}x \\ &= \int \frac{\sec^2 x + \sec x \tan x}{\sec x + \tan x}\text{ d}x \\ &= \int \frac{\left(\sec x + \tan x\right)^{\prime}}{\sec x + \tan x}\text{ d}x \\ &= \ln \left |\sec x + \tan x \right |+c. \end{align*}$

$\noindent The trick for getting $\int \csc x \text{ d}x = -\ln \left|\csc x + \cot x \right | + c = \ln \left |\csc x - \cot x \right | + c$ is similar. Multiply top and bottom of the integrand by $\csc x + \cot x$ and observe that the numerator becomes the negative derivative of the denominator (or multiply top and bottom by $\csc x - \cot x$ and observe the numerator becomes the derivative of the denominator).$

Thankyou once again Integrand
You legend

Was wondering how I can stop leeching off the BOS community and give something back LOL
Probably not in the maths section, will look forward to helping out the physics and chemistry students tho

(after hscs

)

Paradoxica · May 23, 2016

$u = \frac{1}{x}$

This will convert the integral into a standard form (look in your booklet bro)

The answer very rapidly comes out to be

$\mathcal{C}-\frac{1}{a} \log{\left(\frac{a}{x} + \sqrt{\frac{a^2}{x^2}+1}\right)}$

InteGrand · May 23, 2016

SammyT123 said:
Ohh ok - Will make sure to remember that
Teacher told me its not a standard integral, so its best to use t-formula
If you dont mind, can you show me how I can finish off the solution from $\\ I = \frac{1}{a}ln(tan\frac{\theta}{2}) + C$
Given t=tan(theta/2)

(just curious thats all)

$\noindent To get the answer back in terms of $x$, we can use the identity above of $\left |\tan\frac{\theta}{2} \right |=\sqrt{\frac{1-\cos \theta}{1+\cos \theta}}$. Then use the following fact: if $\cos \theta > 0$ (which we can assume because we can impose $-\frac{\pi}{2}<\theta < \frac{\pi}{2}$ when doing the original substitution), we have $\cos \theta = \frac{1}{\sqrt{1+\tan^2 \theta}}$ (because the thing under the square root is just $\sec^2 \theta$). Since $\tan \theta = \frac{x}{a}$, this gives $\cos \theta = \frac{1}{\sqrt{1+\frac{x^2}{a^2}}}=\frac{a}{\sqrt{a^2 + x^2}},a>0$.$

Paradoxica · May 23, 2016

InteGrand said:
$\noindent To get the answer back in terms of $x$, we can use the identity above of $\left |\tan\frac{\theta}{2} \right |=\sqrt{\frac{1-\cos \theta}{1+\cos \theta}}$. Then use the following fact: if $\cos \theta > 0$ (which we can assume because we can impose $-\frac{\pi}{2}<\theta < \frac{\pi}{2}$ when doing the original substitution), we have $\cos \theta = \frac{1}{\sqrt{1+\tan^2 \theta}}$ (because the thing under the square root is just $\sec^2 \theta$). Since $\tan \theta = \frac{x}{a}$, this gives $\cos \theta = \frac{1}{\sqrt{1+\frac{x^2}{a^2}}}=\frac{a}{\sqrt{a^2 + x^2}},a>0$.$

I personally think this situation is simply not necessary.

t-formulae are a path to disaster.

Paradoxica · May 23, 2016

leehuan said:
T-formulae does allow the integral of cosec to come out nice and tidy, but there is no need for it at all

Refer to my previous post.

Also half the time a trig substitution is unnecessary for indefinite integrals.

But for definite integrals they are very slick.

leehuan · May 23, 2016

Paradoxica said:
Refer to my previous post.

Also half the time a trig substitution is unnecessary for indefinite integrals.

But for definite integrals they are very slick.

It's only tidy, if you have cosec by itself. (And it's still not necessary either, but it makes the final answer tidy)

Do we need to memorise this identity? (3 Viewers)

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