Inverse Trigonometric Functions

[premier]

I'm having problems with finding the integrals of inverse trig functions. For example, I'm not sure why or what to substitute for U. In this example

Integral of 1 / x*sqrt [x^2 - 25]
The book substituted x =5u and dx = 5du.

Another example
Integral of 1 / sqrt [1-4x^2]
u =2x and dx = 1/2 du

I don't understand the logic behind the substitution, x=5u and u = 2x. How do you choose what to substitute? Why is it being substituted as 5u and 2x instead of something else? I hope that my problem is comprehensible. Thanks for the help

 

[soroban]

Hello, premier!

I'm having problems with finding the integrals of inverse trig functions.
For example, I'm not sure why or what to substitute for U. In this example

Integral of 1 / x*sqrt [x^2 - 25]
The book substituted x = 5u and dx = 5 du.

Another example
Integral of 1 / sqrt [1 - 4x^2]
u = 2x and dx = 1/2 du

I don't understand the logic behind the substitution, x = 5u and u = 2x.
How do you choose what to substitute?
Why is it being substituted as 5u and 2x instead of something else?

It looks like you have a textbook that uses "1" in everything . . . (How I hate those!)

In the first problem, the denominator has: . x (x² - 25)^1/2

It would fit your arcsec formula, except the formula has: . u (u² - 1)^1/2

To make our problem fit the formula, let: .x = 5u ... dx = 5 du
. . Then: . x² - 25 . = . (5u)² - 25 . = . 25u² - 25 . = . 25(u² - 1)
. . With the square root, we have: . 5(u² - 1)^1/2

The entire denominator is: . x(x² - 25)^1/2 . = . 5u·5(u² - 1)^1/2 . = . 25u(u² - 1)^1/2

. . . . . . . . . . . . . . . . . . . . 5 du . . . . . . . . 1 . . . . . . . . .du
The integral is: . [int] ----------------- . = . --- [int] ----------------
. . . . . . . . . . . . . . . . . 25u(u² -1)^1/2 . . . . 5 . . . . . u(u² - 1)^1/2

NOW it fits the arcsecant formula . . . you can do the integration.
. . . (Don't forget to back-substitute.)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

[rant]
I would give you a much simpler formula for these problems . . .
. . but every time I do, SOMEONE counters with
. . "I'd rather have my students understand the reasoning
. . instead of just memorizing formulas!"
as if LEARNING a few formulas is a terrible thing.

If "Learning = Memorizing = Evil", then why have formulas at all?

If you want the derivative of f(x) = x⁴, use the differerence quotient.
If you want the distance between two points, make a right triangle and use Pythagorus.
If you want the equation of a parabola, derive it from its focus and directrix.
If you want 3 x 7, find 7 + 7 + 7.
. . (God forbid we should memorize anything!)


Okay, one more time . . . I do NOT advocate blind memorization, got it?
I taught math for thirty-eight years (successfully, I'd say) with clear explanations and derivations.
When I derived a formula for my students, I'd give them the best formula available, not one that
requires additional (and tedious) adjustments. .If such adjustments are necessary, I would
<u>incorporate</u> them into the formula . . . and we'd use the generalized formula.

So why all the flak? . Since you must MEMORIZE the arcsecant formula anyway,
why not learn one which eliminates all the work I did above? . (All of it!)

I'll let my opponents explain the second problem . . . they deserve it.
[/rant]

 

[tkhunny]

1 / sqrt [1-4x^2]

Just for the record, I am NOT one of your opponents on this matter. I find the "1"-thing entirely misdirected. The crux of the problem is recognizing the appropriate trig substitution. Fooling with the constant hardly seems a worthwhile hint.

I note these relationships:

This is where life begins.
1) cos²(u) + sin²(u) = 1

Subtract sin²(u) from 1)

2) cos²(u) = 1 - sin²(u)

This gives a way to substitute for Number² - Function². Substitute some version of sin(u) to get some version of cos(u).

Divide 1) by cos²(u)

3) 1 + tan²(u) = sec²(u)

This gives a way to substitute for Number² + Function². Substitute some version of tan(u) to get some version of sec(u).

Subtract 1 from 3)

4) tan²(u) = sec²(u) - 1

This gives a way to substitute for Function² - Number². Substitute some version of sec(u) to get some version of tan(u).

1 / sqrt [1-4x^2]

Number² - Function²
1 - 4x² = 1² - (2*x)²
2*x = sin(u)
2*dx = cos(u)*du

Deal with the constants.
Don't forget to substitute back to 'x'.
You are done.

If you are not going to memorize them, you MUST KNOW how to come up with them.