Integration by trignometric substitution?
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.
Integration using substitution u = sec x?
Let u = sec(x) dx <==> du = sec(x)tan(x) dx. The integral becomes: ∫ sec^3(x)*tan^3(x) dx = ∫ sec^2(x)*tan^2(x) [sec(x)tan(x) dx] = ∫ sec^2(x) * [sec^2(x) - 1] [sec(x)tan(x) dx], since tan^2(x) = sec^2(x) - 1. Applying our substitutions now yields: ∫ sec^2(x) * [sec^2(x) - 1] [sec(x)tan(x) dx] = ∫ u^2*(u^2 - 1) du = ∫ (u^4 - u^2) du = (1/5)u^5 - (1/3)u^3 + C. Back-substituting u = sec(x) yields: ∫ sec^3(x)*tan^3(x) dx = (1/5)sec^5(x) - (1/3)sec^3(x) + C. I hope this helps!
[DESPERATE] Integrate sqrt (4-x²) with trig substitution u = 2sin x?
Let x = 2 sin θ. Then dx = 2 cos θ dθ. Thus, plugging in for x and for dx in the integral, we obtain ∫√(4 - x^2) dx = ∫√(4 - (2 sin θ)^2) (2 cos θ dθ) = ∫√(4 - 4 sin^2 θ)(2 cos θ dθ) = ∫√(4(1 - sin^2 θ))(2 cos θ dθ) = ∫√(4 cos^2 θ)(2 cos θ dθ) = ∫(2 cos θ)(2 cos θ dθ) = ∫ 4 cos^2 θ dθ. I think that you are neglecting the "dx". It's an important part of the integral, and if you make a change of variable from x to θ, then your new integral must possess a dθ; to convert the dx to dθ, you must express dx in terms of dθ and substitute in. --- Here is how to convert sin 2θ back to x-values. Since x = 2 sin θ, then x/2 = sin θ. Now, using x/2 = sin θ, we can draw out a right triangle with one angle θ, the opposite side of length x, and the hypotenuse of length 2. By the Pythagorean Theorem, the side adjacent to θ is of length √(4 - x^2). Thus: sin (2θ) = 2 sin (θ) cos (θ) [by a trigonometric identity] = 2 * (opposite / hypotenuse) * (adjacent / hypotenuse) [by definition of sine and cosine] = 2 * (x / 2) * (√(4 - x^2) / 2) [by the computations above involving the right triangle] = (x/2) √(4 - x^2). [simplification]. Thus, 2θ + sin (2θ) + C = sin^-1 (x/2) + (x/2) √(4 - x^2) + C.
Question on reciprocal trig functions?
ok sounds good to me, its true that is how I first learned of those functions. I suppose its just an easy way of remembering them. The line of reasoning I was thinking of was that: secx/cosecx = tanx which sort of makes sense considering sin/cos is tan. I thought there may have been some naming relationship as sinh/cosh is also tanh, so maybe if you say a function over co-function is tan but it may not be like that. thanks anyway :)
Question about trig sub (Calc)?
I'm a bit stuck on this problem. I know that I have to substitute for tanø as it is added so I get: 5/2sec^2ø/(5/2tanø*5/2secø) then I get (5/2)secø/tanø which simplifies to (5/2)cscø. I then take the integral of that function and convert from ø->x. I don't think i'm getting the right answer though. What is the right way to do this problem?
How do I know which Integration technique to use when solving a problem?
I agree with Kyle Gray that more than anything it takes a ton of practice. The more you practice, the more you’ll start to recognize different kinds of integrals and know right away which method to use.However, I would recommend considering integration methods in the order below. If you’re not immediately sure which method to use, I like running through this checklist because these methods are more or less listed from easiest and quickest to most difficult and longest. In other words, if you really don’t know where to start, why consider the much more complicated trigonometric substitution when a simple u-substitution might work?I’d suggest considering these methods, in this order:Direct integration. If it’s something like a polynomial function that can be integrated with power rule, you’ll be able to integrate directly.U-substitution. Look to see if the integrand contains both a function and its derivative.Integration by parts. The integrand will be one of these:The product of an exponential function ([math]e^x[/math]) and a trig function ([math]\sin{x}[/math])The product of a power function ([math]x^2[/math]) and a trig function ([math]\sin{x}[/math])The product of an exponential function ([math]e^x[/math]) and a power function ([math]x^2[/math])Partial fractions. The integrand will be a rational function (a fraction in which the numerator and denominator are both polynomial functions). You’ll need to factor the denominator.Can you integrate using a trig integral, inverse trig integral, hyperbolic integral, or inverse hyperbolic integral formula? For example, there’s a specific, but simple way to integrate these kinds of functions:[math]\sin^m{x}\cos^n{x}[/math] where [math]m[/math] is odd[math]\sin^m{x}\cos^n{x}[/math] where [math]n[/math] is odd[math]\sin^m{x}\cos^n{x}[/math] where [math]m[/math] and [math]n[/math] are even[math]\sin{(mx)}\cos{(nx)}[/math][math]\sin{(mx)}\sin{(nx)}[/math][math]\cos{(mx)}\cos{(nx)}[/math]Trigonometric substitution. The integrand will contain [math]x^2-a^2[/math], [math]x^2+a^2[/math], or [math]a^2-x^2[/math], where [math]x[/math] is a variable and [math]a[/math] is a constant.Improper integrals. The upper or lower limit of integration will be infinite, or there will be an infinite discontinuity at one end of the interval or inside the interval.Good luck!
What is the difference between trig sub integrals and inverse trig integrals?
There isn't really a difference. It's not unusual to have more than one approach available to integrate a particular expression. If you happen to recognize a function as being the derivative of some well known function, such as knowing that d/dx sin^(-1)(x/a) = 1/√(a² - x²), just use that knowledge. If you don't know (or remember) this as a general rule, use some technique such as trig substitution. It's probably easy to recognize ∫ dx/√(1 - x²) as being arcsin(x) + C. A similar integral that's not so readily remembered is ∫ √(1 - x²) dx which is ½√(1 - x²) + ½arcsin(x) + C. Fortunately you can obtain either result by trig substitution. Some people just remember that ∫ ln(x)dx = xln(x) - x + C, others get there by integration by parts. All of the inverse trig integration "formulas" can be derived from trig substitution.