A Smart Way to Avoid a Lot of Inverse Hassle?

gerardhoyle

Hi everybody :approve:

I'm having a lot of trouble remembering all of the formulas for differentiating inverse trig functions, for integrating trig functions, or integrating inverse trig functions, or integrating hyperbolic trig functions, or integrating inverse hyperbolic trig functions.

Luckily, in differentiating inverse trig functions I can invert the equation & use implicit differentiation & rederive all 6 trig functions all just by knowing how to differentiate trig functions.

That to me is a lot smarter & better than cheating by memorizing formulas as there is simple logic involved.

However, I've had so much trouble trying to deal with the rest of these functions in a logical way & every source I check tells me to just use the formula's I've memorized.

Thomas Calculus
Stewart Calculus
Apostol Calculus
Calculus Made Easy
The Calculus Lifesaver
Loads of youtube videos
Loads of internet pdf's and sites

I've had some success with a particular triangle method, i.e. look for the sum of two perfect squares in a denominator under a square root in the denominator, that means you can draw a triangle & use a very similar logic to rederive your way to the answer according to the situation.

Namely the logic purported in this and this.

However, what about all these seperate cases,

inverse trig functions,
hyperbolic trig
inverse hyperbolic trig

Have all of you just memorized the formula's to just plug in every time you recognise the shape of the equation?

If not I'd be so happy to hear of sources to where you learned this smart way that I seem to be missing. I've had so much trouble with calculus in that I have to ignore my books & search out the real secret behind each method in calculus, for instance;

Newtons method is just an algebraic manipulation of the point-slope form of an equation, two seconds to rederive it.
Linear Approximation is another manipulation of this simple equation in a slightly different way.
Mean Value Theorem is just Rolle's Theorem slanted.
Integral MVT is just standard averaging calc'd up.
All of that shells and washers stuff is just geometry & a flavour of calc.

There just has to be some way to look at trig integrals in a similarly logical way.

gerardhoyle

Here is an example of the method I use that works pretty well, it's got a lot of kinks I'm trying to iron out, and you'll see what I mean.



[latex]\int \frac{dx}{\sqrt{e^{2x} - 6}} [/latex]

by the picture,

[latex] tan \theta = \frac{\sqrt{6}}{\sqrt{e^{2x} - 6}} [/latex]

and,

[latex] \frac{1}{\sqrt{6}}tan \theta = \frac{1}{\sqrt{e^{2x} - 6}} [/latex]

so the integral becomes;

[latex] \int \frac{dx}{\sqrt{e^{2x} - 6}} =\int \frac{1}{\sqrt{6}}tan \theta dx [/latex]

Now to calculate dx, (this gets a bit nuts);

[latex] u = e^x [/latex]

[latex] du = e^x dx [/latex]

[latex] dx = \frac{du}{e^x} = \frac{du}{u} [/latex]

so, the integral is NOW;

[latex] \int \frac{1}{\sqrt{6}} \ tan \theta \ \frac{du}{u} [/latex]


We first get rid of the [latex]u[/latex] by using the fact that,

[latex] sin \theta = \frac{\sqrt{6}}{u} [/latex]

[latex] \frac{1}{\sqrt{6}} \ sin \theta = \frac{1}{u} [/latex]

and this gets our integral even closer;

[latex] \int \frac{1}{\sqrt{6}} \ tan \theta \ \frac{1}{\sqrt{6}} \ sin \theta du [/latex]

cleaning it up;

[latex] \int \frac{1}{6} \ tan \theta \ sin \theta du [/latex]

All thats left to get rid of is [latex]du[/latex],

Because

[latex] sin \theta = \frac{\sqrt{6}}{e^x} [/latex]

and,

[latex] e^x = u = \frac{\sqrt{6}}{sin \theta} = \sqrt{6} \ csc \theta [/latex]

we see that;

[latex] u = \sqrt{6} \ csc \theta [/latex]

so,

[latex] du = - \sqrt{6} \ csc \theta \ cot \theta \ d \theta [/latex]

and the original integral finally becomes:

[latex] \int \frac{1}{6} \ tan \theta \ sin \theta \ du = \int - \frac{1}{6} \ tan \theta \ sin \theta \ \sqrt{6} \ csc \theta \ cot \theta \ d \theta [/latex]

Cleaning it up;

[latex] \int - \frac{1}{\sqrt{6}} \ tan \theta \ sin \theta \ csc \theta \ cot \theta \ d \theta [/latex]

and this reduces to;

[latex] - \frac{1}{\sqrt{6}} \int d \theta [/latex]


[latex] - \frac{1}{\sqrt{6}} \ \theta + C [/latex]

to get theta I'll use;

[latex] sin \theta = \frac{\sqrt{6}}{e^x} [/latex]

[latex] \theta = arcsin ( \frac{\sqrt{6}}{e^x}) [/latex]

so the final answer I get is;

[latex] - \frac{1}{\sqrt{6}} \ arcsin ( \frac{\sqrt{6}}{e^x}) + C [/latex]

ergo;

[latex] \int \frac{dx}{\sqrt{e^{2x} - 6}} dx = - \frac{1}{\sqrt{6}} arcsin( \frac{\sqrt{6}}{e^x}) + C [/latex]

But my book says the answer is;

[latex] \int \frac{dx}{\sqrt{e^{2x} - 6}} = \frac{1}{\sqrt{6}} arcsec( \frac{e^x}{\sqrt{6}}) + C [/latex]

I am supposed to blatantly copy backwards using the method my book
teaches me to do but there is no real magic in this method, it is contrived magic with no thinking and no nothing to copying.

There must of been some slip-up but I haven't seen it

Could you correct me in this method, it is very general insofar as I can tell,
there might be some completing the square in the denominator to get to this
form but it's very general insofar as I can tell.

What did I do wrong here?

I don't think I'm differentiating right as I keep getting weird answers.

But, the method I've used, where is it wrong?

Fremen

You might have better luck if you make the u substitution before drawing the triangle, so that u^2 is on your hypotenuse, u root six is on your opposite and sqrt(u^4 - 6 u^2) is on your adjacent. That would at least cut down on the algebra.

You have a very good approach to this stuff. Figuring things like this out for yourself is absolutely the best way to get good.

gerardhoyle

Fremen wrote: »

You might have better luck if you make the u substitution before drawing the triangle, so that u^2 is on your hypotenuse, u root six is on your opposite and sqrt(u^4 - 6 u^2) is on your adjacent. That would at least cut down on the algebra.

You have a very good approach to this stuff. Figuring things like this out for yourself is absolutely the best way to get good.

Great, I'll try and do the u substitutions earlier, thanx

Well, I'm having a really hard time with this one problem.

This method usually works like a charm for every problem I've done. I'm getting correct answers & completely avoiding all of that memorization of inverse trig functions.

I'm going to think about hyperbolic substitutions when I figure out what's up with this one equation.

This is the books answer differentiated;

[latex] y = \frac{1}{\sqrt{6}} \cdot arcsec(\frac{e^x}{\sqrt{6}}) + C [/latex]

[latex] sec(\sqrt{6}y) = sec[arcsec(\frac{e^x}{\sqrt{6}})] + sec(C) [/latex]

[latex] sec(\sqrt{6}y) = \frac{e^x}{\sqrt{6}} + sec(C) [/latex]

[latex] sec(\sqrt{6}y) tan(\sqrt{6}y) \frac{dy}{dx} = \frac{e^x}{\sqrt{6}} [/latex]

[latex] \frac{dy}{dx} = \frac{e^x}{\sqrt{6}} \cdot \frac{1}{ sec(\sqrt{6}y) tan(\sqrt{6}y) } [/latex]

[latex] \frac{dy}{dx} = \frac{\sqrt{6}}{e^x} \cdot \frac{e^x}{\sqrt{6}} \cdot \frac{1}{ tan(\sqrt{6}y) } [/latex]

[latex] \frac{dy}{dx} = \frac{1}{ tan(\sqrt{6}y) } [/latex]

[latex] tan(\sqrt{6}y) = \sqrt{sec^2(\sqrt{6}y) - 1} [/latex]

[latex] \frac{dy}{dx} = \frac{1}{ \sqrt{sec^2(\sqrt{6}y) - 1} } [/latex]

[latex] \frac{dy}{dx} = \frac{1}{ \sqrt{\frac{e^{2x}}{6} - 1} } [/latex]

I'm not sure exactly how to legally factor that in the denominator to get it into the form the original integral is in but I'm 99.9999% sure it is legal. Could you help?

This is my answer differentiated;

[latex] y = - \frac{1}{\sqrt{6}} \cdot arcsin ( \frac{\sqrt{6}}{e^x}) + C [/latex]

[latex] sin(\sqrt{6}y) = - \frac{\sqrt{6}}{e^x} + sin(C) [/latex]

[latex] cos(\sqrt{6}y) \cdot \frac{dy}{dx} = \frac{\sqrt{6}}{e^x} [/latex]

[latex] \frac{dy}{dx} = \frac{\sqrt{6}}{e^x} \cdot \frac{1}{cos(\sqrt{6}y)} [/latex]

[latex] \frac{dy}{dx} = \frac{\sqrt{6}}{e^x} \cdot \frac{1}{\sqrt{1 - sin^2 ( \sqrt{6} y)} [/latex]

[latex] \frac{dy}{dx} = \frac{\sqrt{6}}{e^x} \frac{1}{ \sqrt{1 - \frac{6}{e^{2x}} [/latex]

Again, I have a feeling that this can be factored into the correct form
as the original integral but I don't see how to do it legally.

Edit: The last two terms wont show in latex for me so in case they
are just the square root of 1 - sin^(root6 y) and then converted into
6 over e^2x

However, there is the case of that extra term sticking out like a sore thumb!

If this answer can be factored then I've gotten it amost right, apart from that term which I maybe missed

I have a feeling this is correct in a similar way that sinθ = -cosθ' what with the argument inside the arcsin and the arcsecant being flipped as well as the inclusion of the minus sign, would that be correct even in arcsin land & is there a surefire way to discern that naturally?

Fremen

gerardhoyle wrote: »

[latex] \frac{dy}{dx} = \frac{1}{ \sqrt{\frac{e^{2x}}{6} - 1} } [/latex]

Multiplying above and below by root six should give you back the original form. Is that what you were asking?

Not sure if this will be helpful, but you can work out from the wikipedia page on inverse trig functions that

arcsec(x) = pi/2 - arcsin(1/x).