Do we need to memorise this identity? (1 Viewer)

SammyT123 · May 23, 2016

Thanks paradoxica

I personally prefer the trig substitution

But jeez that method is also very quick

SammyT123 · May 23, 2016

Are the t-formula results ever 'unavoidable'

(I know that theoretically they are obviously unavoidable, but sitting in the middle of a 4U hsc exam is it still feasible to avoid them?)

SammyT123 · May 23, 2016

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InteGrand · May 23, 2016

SammyT123 said:
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$\noindent I think you mis-evaluated the integral. Note that if $a>0$, then$

$\begin{align*} \int \frac{1}{\sqrt{a^2 u^2 -1}}\text{ d}u &= \frac{1}{a}\int \frac{1}{\sqrt{u^2 -\frac{1}{a^2}}}\text{ d}u \\ &= \frac{1}{a} \ln \left(u+\sqrt{u^2 -\frac{1}{a^2}}\right). \end{align*}$

$\noindent Here we made use of the standard integral $\int \frac{1}{\sqrt{\chi^2 - \alpha^2}}\text{ d}\chi = \ln \left(\chi +\sqrt{\chi^2 - \alpha^2} \right)$.$

leehuan · May 23, 2016

SammyT123 said:
Are the t-formula results ever 'unavoidable'

(I know that theoretically they are obviously unavoidable, but sitting in the middle of a 4U hsc exam is it still feasible to avoid them?)

Something like this is just dodge

$\int_0^\frac{\pi}{6}{\frac{dx}{3+4\sin{x}+5\cos{x}}}$

SammyT123 · May 23, 2016

InteGrand said:
$\noindent I think you mis-evaluated the integral. Note that if $a>0$, then$

$\begin{align*} \int \frac{1}{\sqrt{a^2 u^2 -1}}\text{ d}u &= \frac{1}{a}\int \frac{1}{\sqrt{u^2 -\frac{1}{a^2}}}\text{ d}u \\ &= \frac{1}{a} \ln \left(u+\sqrt{u^2 -\frac{1}{a^2}}\right). \end{align*}$

$\noindent Here we made use of the standard integral $\int \frac{1}{\sqrt{\chi^2 - \alpha^2}}\text{ d}\chi = \ln \left(\chi +\sqrt{\chi^2 - \alpha^2} \right)$.$

Thanks
Im so out of it today - Im off for now

SammyT123 · May 23, 2016

leehuan said:
Something like this is just dodge

$\int_0^\frac{\pi}{6}{\frac{dx}{3+4\sin{x}+5\cos{x}}}$

Haha thats exactly the type of question I meant

Paradoxica · May 23, 2016

SammyT123 said:
Are the t-formula results ever 'unavoidable'

(I know that theoretically they are obviously unavoidable, but sitting in the middle of a 4U hsc exam is it still feasible to avoid them?)

Sit next to me and ...

well you know how the rest goes.

SammyT123 · May 23, 2016

Paradoxica said:
Sit next to me and ...

well you know how the rest goes.

LOL If only I could

Paradoxica · May 23, 2016

leehuan said:
Something like this is just dodge

$\int_0^\frac{\pi}{6}{\frac{dx}{3+4\sin{x}+5\cos{x}}}$

Pretty sure I have a formula for that integral form.....

Paradoxica · May 23, 2016

Also one of the things to keep in mind for the tangent half angle substitution is that you have to take care when integrating across a singularity....

Because that can give different results if you ignore the singularity versus you splitting and taking limits.

Paradoxica · May 23, 2016

leehuan said:
Something like this is just dodge

$\int_0^\frac{\pi}{6}{\frac{dx}{3+4\sin{x}+5\cos{x}}}$

Also

1. Yuck. The numbers aren't even nice. Would have been lovely to have 3sinx + 4cosx but nooooooooo you didn't give me a correctly ordered pythagorean triple

2. t-formula is messy answer anyway, since it involves a fair bit of surds

3. *flies away*

SammyT123 · May 23, 2016

Paradoxica said:
Also

1. Yuck. The numbers aren't even nice. Would have been lovely to have 3sinx + 4cosx but nooooooooo you didn't give me a correctly ordered pythagorean triple

2. t-formula is messy answer anyway, since it involves a fair bit of surds

3. *flies away*

Went in balls deep
Didn't seem too bad unless I screwed up completely LOL

Paradoxica · May 23, 2016

SammyT123 said:
Went in balls deep
Didn't seem too bad unless I screwed up completely LOL

I had √2, √3 appear in the answer.

If I see more than one non-square number under a square root, I will immediately back out and look to see if there is a better path to the answer.

Otherwise, proceed through the flames of hell.

SammyT123 · May 23, 2016

Paradoxica said:
I had √2, √3 appear in the answer.

If I see more than one non-square number under a square root, I will immediately back out and look to see if there is a better path to the answer.

Otherwise, proceed through the flames of hell.

What did I do wrong this time
Rip

Paradoxica · May 23, 2016

SammyT123 said:
What did I do wrong this time
Rip

It's a definite integral... :L

Which also reminded me I did the borders incorrectly.

Time to redo this.

Paradoxica · May 23, 2016

Paradoxica said:
It's a definite integral... :L

Which also reminded me I did the borders incorrectly.

Time to redo this.

Not even half as bad as I thought it was, but still not particularly nice.

Drongoski · May 24, 2016

I recall doing a similar integral a few days ago, from Bob Aus's book.

Using the usual t-formulae:

$\int ^{\frac {\pi}{2}} _0 \frac {d\theta}{5cos \theta + 12sin \theta + 13}\\ \\ = \int ^1 _ 0 \frac {\frac {2}{1+t^2} dt}{\frac {5(1-t^2)}{1+t^2} + \frac {12 \times 2t}{1+t^2} + \frac {13(1+t^2)}{1+t^2}}\\ \\ = \cdots \cdots \\ \\ = \int ^1 _0 (2t+3)^{-2} dt \\ \\ = \cdots \cdots \\ \\ = \frac {1}{15}$

Paradoxica · May 24, 2016

Drongoski said:
I recall doing a similar integral a few days ago, from Bob Aus's book.

Using the usual t-formulae:

$\int ^{\frac {\pi}{2}} _0 \frac {d\theta}{5\cos \theta + 12\sin \theta + 13}\\ \\ = \int ^1 _ 0 \frac {\frac {2}{1+t^2} dt}{\frac {5(1-t^2)}{1+t^2} + \frac {12 \times 2t}{1+t^2} + \frac {13(1+t^2)}{1+t^2}}\\ \\ = \cdots \cdots \\ \\ = \int ^1 _0 (2t+3)^{-2} dt \\ \\ = \cdots \cdots \\ \\ = \frac {1}{15}$

The Auxiliary Transformation and the Tangent Half Angle Substitution will be guaranteed to be nice for (asinx + bcosx + c)^-1 IFF (a,b,c) is a Pythagorean Triple.

Drongoski · May 24, 2016

Paradoxica said:
The Auxiliary Transformation and the Tangent Half Angle Substitution will be guaranteed to be nice for (asinx + bcosx + c)^-1 IFF (a,b,c) is a Pythagorean Triple.

That's interesting. I have not investigated, but I noticed that in both integrals, {3,4,5} and {5,12,13} are Pythagorean Triples.

Do we need to memorise this identity? (1 Viewer)

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