Antidifferentiate this (Inverse trigonometry)?

Question

Hello, Question reads:

Antidifferentiate this (an Inverse trigonometry question):

dx / square root of (9-3x^2)

(the "/" above means "over" as in the fraction sign)

Any pointers? thanks

Answer 1

There is a difference between differential calculus and trigonometry (and you can combine both).

The "anti-differentiation" is akin to the integral.

When you have a function, the differential is like finding the slope of the curve. The integral is like finding the area under the curve.

The anti-differential means: what was the function for which the differential is dx/SQRT(9-3x^2).

The "inverse trigonometry" in the question is a hint that dx/SQRT(9-3x^2) could the derivative of an inverse trigonometric function (probably the inverse of a sine = arc-sine).

The arcsin of k is the angle T for which the sine is k:
sin(T) = k
arcsin(k) = T

d(arcsin(k)) = dk/SQRT(1-k^2)

Then you have to play with k, making it a function of x; you may also have to find a coefficient to make the values come out to 9 - 3x^2. From that point on, it could be done with trial and error.

----

For those who wonder: yes, I stopped there because I was stuck. I had to grab a pencil and work it out the old-fashioned way.

I took
dx/SQRT(9-3x^2) and tried to make it fit the pattern
dk/SQRT(1-k^2).

I used
dx/SQRT{9[1 - (x^2)/3]}
dx/{3*SQRT[1 - (x^2)/3]}
(1/3)* dx/SQRT[1 - (x^2)/3]

then I make k = x/√3 (x divided by SQRT(3))
I also need the fact that dk = dx/√3
If k = (1/√3)x, then dk = d[(1/√3)x] = (1/√3)d(x) = (1/√3)dx
From that, I find that dx = √3*dk. I substitute for all x's:

(1/3)* dx/SQRT[1 - (x^2)/3]
(1/3)* √3 dk / SQRT(1 - k^2)

Now I can solve it:

If
d(arcsin(k)) = dk/SQRT(1-k^2)
then
(√3 / 3) * d(arcsin(k)) = (√3 / 3) * dk/SQRT(1-k^2)
d[(√3 / 3) * (arcsin(k))] = √3 dk/[3*SQRT(1-k^2)]
d[(√3 / 3) * (arcsin(k))] = √3 dk/SQRT[9*(1-k^2)]
d[(√3 / 3) * (arcsin(k))] = √3 dk/SQRT(9 - 9*k^2)

replace k by x/SQRT(3) (then k^2 = (1/3)x^2)

d[(√3 / 3) * (arcsin(x/√3))] = √3 dk/SQRT[9 - (9/3)x^2]
d[(√3 / 3) * (arcsin(x/√3))] = √3 dk/SQRT(9 - 3x^2)

Then replace dk with dx/√3 (then √3 dk becomes simply dx)

d[(√3 / 3) * (arcsin(x/√3))] = dx/SQRT(9 - 3x^2)

"The differential of √3 / 3) * (arcsin(x/√3)) is equal to dx/SQRT(9 - 3x^2)"

Therefore, the antidifferential of dx/SQRT(9 - 3x^2) is

(√3 / 3) * (arcsin(x/√3))

Answer 2

ok...write 8 - x^2 - 2*x as 8 +a million -a million - x^2 - 2*x ...i.e. you upload and subtract a million, so now the expression will become 9 - (a million+x)^2. Now exchange a million+x = 3 sin u.s., dx = 3 cos u du The necessary will become: a million/sqrt(8 - x^2 - 2*x) dx = a million/sqrt(9 - (a million+x)^2) dx = (a million/sqrt(9 - 9 sin^2 u) ) 3 cos u du = du = u = arcsin (a million+x)/3) arcsin is inverse sin.....

Answer 3

Use trig substitution:

Let x=(3/sqrt(3))sin(v)

Then dx=(3/sqrt(3))cos(v)dv

And we get the integrand to be

[1/sqrt(9-9sin^2(v))][(3/sqrt(3))cos(v)dv]

=1/sqrt(3) dv

=1/sqrt(3)v+C

=(arcsin(x sqrt(3)/3))/sqrt(3)+C

Note: It's the same as Vince's answer, slightly different form.

Answer 4

The key for inverse trig integrals is to do the right substitution. For this case, you have a - bx^2, which is an inverse sin substitution (a + bx^2 would be an inverse tan substitution). Here you want sqrt(3)*x = 3sin(theta), such that sqrt(3)dx = 3cos(theta)d(theta).

Integral becomes:
(3/sqrt(3))cos(theta)d(theta) / sqrt(9-9sin^2(theta))
=sqrt(3)cos(theta)d(theta) / (3*sqrt(1-sin^2(theta)))
=sqrt(3)cos(theta)d(theta) / (3*sqrt(cos^2(theta))
=(sqrt(3)/3)cos(theta) d(theta) / cos(theta)
=(sqrt(3)/3)d(theta)

The integral of which is just (sqrt(3)/3)*theta + c

Then just plug in theta = inv_sin((sqrt(3)/3)*x) to get your answer:

(sqrt(3)/3) inv_sin((sqrt(3)/3)*x) +c