I=∫(x^2)/(4-x^2)^(3/2)
Help please?
Help please?
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Whenever you see something with a denominator that looks something like that, you should think of trig substitution (this allows you to replace a two term denominator of a certain kind to one term only).
let x = 2 sin u, du = 2 cos u
the denominator (4 - x^2)^(3/2) = (4 - 4 sin^2 u)^(3/2) = 4^(3/2) (1 - sin^2 u)^(3/2) = 8 (1 - sin^2 u)^(3/2)
note that cos^2 u + sin^2 u = 1 --> 1 - sin^2 u = cos^u, so
8 (1 - sin^2 u )^(3/2) = 8 (cos^2 u)^(3/2) = 8 cos^3 u
also note that in the numerator we have x^2 = 4 sin^2 u according to our substitution, and dx = 2 cos u du, so the integral in terms of u becomes:
∫(x^2)/(4-x^2)^(3/2) dx
∫ {4 sin^2 u / { 8 cos^3 u} } (2 cos u du)
= ∫ { sin^2 u cos u / {cos^3 u} } du
= ∫ sin^2 u / cos^2 u du
= ∫ tan^2 u du
now, recall that cos^2 u + sin^2 u = 1, dividing by cos^2 u, we get 1 + tan^2 u = sec^2 u --> tan^2 u = sec^2 u - 1
∫ tan^2 u du = ∫ (sec^2 u - 1) du
= ∫sec^2 u du - ∫ du
= tan u - u
It is well-known that d/du tan u = sec^2 u, if you want to show this you can compute the derivative of tan u and verify it really is sec^2 u = 1/cos^2 u
So we have
I = tan u - u
substitute back in for x, recall we let x = 2 sin u, so u = sin^-1 (x/2),
I = tan { sin^-1 (x/2) } - sin^-1 (x/2)
Now you could leave it like this and you would be correct, but the first term looks silly, right? I purposefully wrote it out like that to illustrate how and why people quote results in certain ways. This is right, no mistake about it, but it would be better if we could recast the first term into a more friendly representation, let us take one step back
I = tan u - sin^-1 (x/2) <--- note this is just for illustrated purposes, we want to have a final answer in terms of only one variable (x).
to see an alternative representation for tan u, recall
x = 2 sin u --> sin u = x/2, this is out original substitution, it prescribes u is an angular-like variable, the sine of u gives a ratio of "triangle" side lengths, i.e.
recall tan u = sin u / cos u
we can get an expression for cos u, by noting cos u = adjacent / hypotenuse, sin u = opposite / hypotenuse = x / 2, so
opposite = x
hypotenuse = 2
adjacent = sqrt{2^2 - x} = sqrt{4 - x^2} <-- by Pythagorean theorem
so cos u = adjacent / hypotenuse = sqrt{ 4 - x^2} / 2, sin u = x/2, thus
tan u = sin u / cos u = x / sqrt{ 4 - x^2}
So we may report our final answer as
I = x / sqrt{ 4 - x^2} - sin^-1 (x/2)
let x = 2 sin u, du = 2 cos u
the denominator (4 - x^2)^(3/2) = (4 - 4 sin^2 u)^(3/2) = 4^(3/2) (1 - sin^2 u)^(3/2) = 8 (1 - sin^2 u)^(3/2)
note that cos^2 u + sin^2 u = 1 --> 1 - sin^2 u = cos^u, so
8 (1 - sin^2 u )^(3/2) = 8 (cos^2 u)^(3/2) = 8 cos^3 u
also note that in the numerator we have x^2 = 4 sin^2 u according to our substitution, and dx = 2 cos u du, so the integral in terms of u becomes:
∫(x^2)/(4-x^2)^(3/2) dx
∫ {4 sin^2 u / { 8 cos^3 u} } (2 cos u du)
= ∫ { sin^2 u cos u / {cos^3 u} } du
= ∫ sin^2 u / cos^2 u du
= ∫ tan^2 u du
now, recall that cos^2 u + sin^2 u = 1, dividing by cos^2 u, we get 1 + tan^2 u = sec^2 u --> tan^2 u = sec^2 u - 1
∫ tan^2 u du = ∫ (sec^2 u - 1) du
= ∫sec^2 u du - ∫ du
= tan u - u
It is well-known that d/du tan u = sec^2 u, if you want to show this you can compute the derivative of tan u and verify it really is sec^2 u = 1/cos^2 u
So we have
I = tan u - u
substitute back in for x, recall we let x = 2 sin u, so u = sin^-1 (x/2),
I = tan { sin^-1 (x/2) } - sin^-1 (x/2)
Now you could leave it like this and you would be correct, but the first term looks silly, right? I purposefully wrote it out like that to illustrate how and why people quote results in certain ways. This is right, no mistake about it, but it would be better if we could recast the first term into a more friendly representation, let us take one step back
I = tan u - sin^-1 (x/2) <--- note this is just for illustrated purposes, we want to have a final answer in terms of only one variable (x).
to see an alternative representation for tan u, recall
x = 2 sin u --> sin u = x/2, this is out original substitution, it prescribes u is an angular-like variable, the sine of u gives a ratio of "triangle" side lengths, i.e.
recall tan u = sin u / cos u
we can get an expression for cos u, by noting cos u = adjacent / hypotenuse, sin u = opposite / hypotenuse = x / 2, so
opposite = x
hypotenuse = 2
adjacent = sqrt{2^2 - x} = sqrt{4 - x^2} <-- by Pythagorean theorem
so cos u = adjacent / hypotenuse = sqrt{ 4 - x^2} / 2, sin u = x/2, thus
tan u = sin u / cos u = x / sqrt{ 4 - x^2}
So we may report our final answer as
I = x / sqrt{ 4 - x^2} - sin^-1 (x/2)
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This is a trig substation problem! You're going to have to manipulate the denominator to equal a trig function (Hint it's going to be sine). Good luck!