Integration by trignometric substitution?

Question

How do i do problems like this? some how my teacher substitues in trig functions like tan cot sec ect. into a completley differnt problem.

The problem is

Find and simplify: Integral( x^2/(64-x^2)^(1/2))

oh, and some how the order of the x^2 and the constant number (x^2-12 vs 12-x^2) matters?

Answer 1

In general, if your integral has √(a^2-x^2) you will use x=a*sin z

If your integral has √(a^1+x^2) , you will use x=a*tan z

If your integral has √(x^2 -a^2), you will use x=a*sec z.

Then you have to use the trig identities to simplify your integrand. The reason for using the different trig functions if subtracted in different order is that some trig identities would then give you a negative square root.

Your problem has √(64-x^2) so you will use x=8sin z. Then
64-x^2 = 64 -(8 sin z)^2
= 64 - 64 sin^2 z
= 64 [ 1- sin^2 z]
= 64 cos^2 z

and the square root of that is 8 cos z.

But if you had used x=8 sec z you would get 1-sec^2 z, which is negative: 1 - sec^2 z = - tan^2 z
so you could not do the square root of it.

Back to your problem: you will also have to convert the dx into dz, by doing the derivative of x=8sin z to get dx = 8 cos z dz.

Then your problem will be to integrate
(64 sin^2 z) / 8 cos z * 8 cos z dz
= 64 sin^2 z dz

Personally I would look that one up in a table. Or, you can use half-angle identities to rewite it AGAIN.

Answer 2

Without guarantee, I make it:

32 sin^(-1) (x/8) - 4 x [1 - (x/8)^2]^(1/2).

[Later: O.K., having gone through it twice, it's fine.]

The required substitution is x = 8 sin z.

This is because then, (64 - x^2) becomes 64 (1 - (sin z)^2)
= 64 cos^2 z, whose square root is obviously just 8 cos z.

Also, x = sin z ==> dx = 8 cos z dz, whose "8 cos z" neatly cancels with that previous one.

So the indefinite integral becomes ∫ (64 sin^2 z) dz =
32 ∫(1 - cos 2z) dz

= 32 z - 16 sin 2z = 32 sin^(-1) (x/8) - 32 sin z cos z

= 32 sin^(-1) (x/8) - 32 (x/8) [1 - (x/8)^2]^(1/2)

= 32 sin^(-1) (x/8) - 4 x [1 - (x/8)^2]^(1/2).

Live long and prosper.

LATER NOTES:

1. With regard to your "Additional details": The order of the x^2 doesn't matter, other than that it's customary to express a fraction compactly by writing the numerator first! However, the question of getting the order right in the surd part
--- (64 - x^2)^(1/2) or (x^2 - 64)^(1/2) --- is CRUCIAL!

The original integral is really valid only for x < 8; its solution has a TRIGONOMETRIC part. Had the surd part of the integrand been (x^2 - 64)^(1/2), the corresponding range would be x > 8. The required substitution can then be found by considering the result for hyperbolic functions corresponding to
cos^2 z + sin^z = 1. The hyperbolic function analogue is:

cosh^2 t - sinh^2 t = 1.

In order to get a useful substitution for doing the integral, re-express this identity as

sinh^2 t = cosh^2 t - 1.

This now has the right look to be a candidate to help solve some kind of "sqrt(Y^2- 1)" integration problem

Thus the substitution for the alternative problem would be

x = 8 cosh t, so that √(x^2 - 64) = √[64 (cosh^2 t - 1)] = 8 sinh t.

You can now see how, in the solution to this different problem, the inverse trig. fn. will be replaced by an inverse hyperbolic function, the form of the surd will be reversed, and the domain will be x > 8 instead of x < 8.

2. Unfortunately, 'fred' (5 answers below) made a mistake very early on, starting from the same substitution. He substituted
64 cos^2 z for x^2 instead of the correct 64 sin^2 z. That explains why he obtained a ' + ' sign ' for the surd part of the solution, instead of the correct ' - ' sign. (I'm afraid that as a lifelong mathematician, I can't help reading and editing other people's work.)

Answer 3

Draw a right triangle and label one of the angles theta. (not the right angle).
Label the hypotenuse as 8.
Label the leg next to theta as x
Use the pythagorean theorem to get the other leg
(64-x^2)^(1/2)

Now you can use the trig functions and this triangle to make your substitutions. Look at the triangle and see which trig function gives you the original formula under the integral. Don't forget you need to substitute for dx also. To get dx you find a trig function that gives a very simple formula for x. In this case cos(theta) = x/8. Now take the derivative to get -sin(theta) = dx / 8 or dx = -8 sin(theta).

Once you finish substituting and solving the integral, don't forget to substitute again to have your answer in terms of x. You do this by again looking at the triangle that you drew.

Answer 4

The general form is

∫x²/√(a²-x²) dx = ½ (a² arcsin[x/a] - x√(a²-x²) where |x|<|a|

∫x²/√(64-x²) dx = ½ (64 arcsin[x/8] - x√(64-x²)

∫x²/√(64-x²) dx = 32 arcsin[x/8] - ½x√(64-x²)

Use the substitution u=√(64-x²) to get this.

Edit: If you switch the terms in the √(a²-x²) term (or if |x|>|a|) then you'd have a very different result as the integrand would be a complex function.

Answer 5

You notice that the denominator has a difference of squares.
And you know that 1 - sin^2x = cos^2 x
So try x^2 = 64 sin^2 u
or x = 8 sinu

dx = 8 cosu *du

Now you end up with
64 sin^2 u * 8 cosu *du / (8 cosu)
= 64 sin^2 u* du

Now convert that to cos(2u) and integrate.

Answer 6

this is a very advanced integration problem and to solve it, you should at least be aware of some standard integration formulas such as ∫1/ √(a² - x²) = 1/a arcsin (x/a) + c

but from the looks of it, your teacher has not gone very deep into it, are you sure you wrote the correct question down? this is uni-level calculus.

Answer 7

∫ x².dx/ √[64 - x²]...................Let x = 8sinu then dx/du = 8cosu and so dx = 8cosu du

= ∫ 64cos²u 8cosu du / √[64 - 64sin²u]

= ∫ 64cos²u 8cosu du / √[64cos²u]

= ∫64cos²u 8cosu du / 8cosu

= ∫64cos²u du

= 32∫2cos²u du

= 32∫1 + cos2u du

= 32u + 16sin2u + c

= 32 arcsin(x/8) + 32sinu cosu + c

= 32 arcsin(x/8) + 32. x/8 . √[1 - sin²u] + c

= 32 arcsin(x/8) + 32. x/8 . √[1 - (x/8)²] + c

= 32 arcsin(x/8) + 4 x √[1 - (x/8)²] + c