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Another integral

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tony873004
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I get the right answer, but...

[tex]
\int_{1/2}^{\sqrt 3 /2} {\frac{6}{{\sqrt {1 - t^2 } }}\,dt} [/tex]

[tex] F(x) = 6\,\sin ^{ - 1} x
[/tex]

[tex]
6\sin ^{ - 1} \frac{{\sqrt 3 }}{2} - 6\sin ^{ - 1} \frac{1}{2} = \pi [/tex]

I only know the answer is pi because I plugged it into my calculator and came out with 6.28... - 3.14... = 3.14...

Is there an easier way besides using the calculator to recognize that this equals pi?
 

Answers and Replies

radou
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This is a table integral, unless I'm mistaken.

[tex]\int \frac{dx}{\sqrt{a^2-x^2}}=\arcsin(\frac{x}{a}) + C[/tex]

Edit: this can be shown by using a trig substitution.
 
Dick
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The exact values of the arcsin function come from working out the angle and side relations in a 30-60-90 triangle (ie half of a equilateral). Maybe you could work them out as a way of reviewing precalc!
 
This is a table integral, unless I'm mistaken.

[tex]\int \frac{dx}{\sqrt{a^2-x^2}}=\arcsin(\frac{x}{a}) + C[/tex]

Edit: this can be shown by using a trig substitution.
I personally learned integration without anyone making reference to 'tables of integrals'. And I don't see the point in them, you begin to think integration is a matter of memorizing hundreds of anti-derivatives instead of actually learning how to get there. (granted when you do become proficient, use whatever means necessary)

The exact values of the arcsin function come from working out the angle and side relations in a 30-60-90 triangle (ie half of a equilateral). Maybe you could work them out as a way of reviewing precalc!
Calculus uses radians :P. you should know that sin(pi / 6) = 1 / 2, and therefore you should be able to figure out what arcsin(1 / 2) is (mind you there are two solutions for all 'arc' functions the exception of critical points).
 
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This is a table integral, unless I'm mistaken.

[tex]\int \frac{dx}{\sqrt{a^2-x^2}}=\arcsin(\frac{x}{a}) + C[/tex]

Edit: this can be shown by using a trig substitution.

Let [tex]t=sin(\theta)[/tex] so that [tex]dt = cos(\theta) d\theta[/tex].

Then substitute in, and you should get something like

[tex]\int\frac{dt}{\sqrt{1-t^2}}[/tex]
[tex]=\int \frac{cos(\theta)}{cos(\theta)} d\theta[/tex]
[tex]=\int d\theta[/tex]
[tex]=\theta + C[/tex], C the constant of integration
but since [tex]t=sin(\theta)[/tex], then [tex]\theta = arcsin(t)+C[/tex]

This way you won't have to use the table of integrals.
 
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mind you there are two solutions for all 'arc' functions the exception of critical points.
No there aren't, otherwise they wouldn't be functions. The inverses of sines, cosines are defined on restricted domains in order to make them functions.
 
HallsofIvy
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I get the right answer, but...

[tex]
\int_{1/2}^{\sqrt 3 /2} {\frac{6}{{\sqrt {1 - t^2 } }}\,dt} [/tex]

[tex] F(x) = 6\,\sin ^{ - 1} x
[/tex]

[tex]
6\sin ^{ - 1} \frac{{\sqrt 3 }}{2} - 6\sin ^{ - 1} \frac{1}{2} = \pi [/tex]

I only know the answer is pi because I plugged it into my calculator and came out with 6.28... - 3.14... = 3.14...

Is there an easier way besides using the calculator to recognize that this equals pi?
So the question is NOT about doing the integral, just about finding the exact values of the inverse sine and cosine functions.

You should have already learned some basic values for sine and cosine. As far as [itex]\frac{\sqrt{3}}{2}[/itex] and [itex]\frac{1}{2}[/itex] are concerned, star with an equilateral triangle. Dropping a perpendicular from one vertex to the base gives you two "30-60" right triangles ([itex]\pi/6[/itex] and [itex]\pi/3[/itex] in radians). The hypotenuse of the right triangle is one side of the equilateral triangle, s. The side opposite the [itex]\pi/6[/itex] angle is [itex]\frac{s}{2}[/itex]. Use the Pythagorean theorem to determine that the other side, the altitude of the equilateral triangle, is [itex]\frac{s\sqrt{3}}{2}[/itex]. From that the sine and cosine values follow:
[tex]sin(\frac{\pi}{3})= \frac{\sqrt{3}}{2}[/tex]
and
[tex]sin(\frac{\pi}{6}}= \frac{1}{2}[/tex]

[tex] 6\frac{\pi}{3}- 6\frac{\pi}{6}= 2\pi- \pi= \pi[/itex]
 
Last edited by a moderator:
radou
Homework Helper
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I personally learned integration without anyone making reference to 'tables of integrals'. And I don't see the point in them, you begin to think integration is a matter of memorizing hundreds of anti-derivatives instead of actually learning how to get there. (granted when you do become proficient, use whatever means necessary)
The purpose of 'table integrals' is to make your life easier. Of course, taking into account that you know why the primitives 'look the way they look', i.e. you know how to obtain these functions. If one begins to think about integration as 'a matter of memorizing hundreds of anti derivatives instead of actually learning how to get there', that's just his and noone else's problem then. :smile:
 
AlephZero
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I personally learned integration without anyone making reference to 'tables of integrals'. And I don't see the point in them, you begin to think integration is a matter of memorizing hundreds of anti-derivatives instead of actually learning how to get there.
Unless you work every integral by doing an epsilon-delta proof from first principles, you DO have a table of integrals in your head already. The only difference is, it's probably a shorter table than most published ones.

You might just as well argue that you don't see the point in using a calculator, because you know how to do arithmetic without one. If you need the value of a trig function, no problem - just work it out by hand from the power series...
 

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