Trigonometric Substitutions - help

In summary, the conversation involved a problem with the integral of a trigonometric function, and the conversation delved into various trigonometric substitutions and techniques to solve the problem. The final problem discussed was how to simplify the integral $\int \sqrt{5+4x-x^2}dx$ using a trigonometric substitution.
  • #1
Slimsta
190
0

Homework Statement


[tex]$\int \frac{\sqrt{1-4x^2}dx}{x}$[/tex]


Homework Equations




The Attempt at a Solution


im stuck and i have no idea why I am getting the wrong answer.

let 2x = sin[tex]\phi[/tex]
dx = cos[tex]\phi[/tex] d[tex]\phi[/tex] / 2
[tex]$\int \frac{\sqrt{1-(2x)^2}dx}{x}$[/tex]=[tex]$\int \frac{\sqrt{1-(sin\phi)^2}dx}{x}$[/tex]=[tex]$\int \frac{\sqrt{cos\phi^2}dx}{x}$[/tex]=[tex]$\int \frac{cos\phi dx}{x}$[/tex]=

[tex]$\int \frac{cos\phi^2 d\phi}{2x}$[/tex]= [tex]$ .5\int \frac{cos\phi^2 d\phi}{x}$ =[/tex] [tex]$ .5\int \frac{.5(1+cos(2\phi)) d\phi}{x}$= [/tex]
[tex]$ (1/4) \int 1/x+cos(2\phi)/x d\phi $[/tex]

am i even on the right truck?
 
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  • #2
When you do your trig substitution, replace every x in your original integral. x is still appearing in the denominator.
 
  • #3
[tex] $\int \frac{cos\phi dx}{2sin\phi}$ = [/tex]
[tex] $.5 \int \frac{cos\phi dx}{sin\phi}$ =[/tex]
[tex] $.5 lnsin\phi + C$[/tex]

but i don't think that's right is it..
 
  • #4
Nope. Start again from the beginning, and exchange x and dx for phi and dphi all in one step.
 
  • #5
so i did the same thing over again and i got this:
[tex]$ (1/4) \int 1/sin\phi - sin\phi d\phi $ [/tex]
which =
[tex]$ (1/4)[ \int 1/sin\phi d\phi + cos\phi]$ [/tex]
right? now what..
 
  • #6
dx = (1/2)cos t dt, and x = (1/2) sint -- t being short for theta

The dx in the numerator and the x in the denominator cause the (1/2)'s to cancel.

So you should have
[tex]\int \left[\frac{1}{sin t} - sin t\right]dt[/tex]
[tex]=\int \frac{1}{sin t}dt - \int sin t dt[/tex]

For the first you need the antiderivative of csc t. The second is easy.

Does you book show you a trick for antidifferentiating sec t? It involves multiplying by 1 in the form of sec t + tan t over itself. There's a similar trick for working with the integral of csc t.
 
  • #7
olympics distracted me for a while lol

yeah i got that antiderivative after googling it hhh.. my book didnt say anything about it.

now I am stuck on this question:
[tex]$\int \frac{dx}{\sqrt{16-x^2}}$[/tex]

i let x = 4sint
dx= 4cost dt

so
[tex]$\int \frac{dx}{\sqrt{16-16sin^2}}$[/tex] =[tex]$\int \frac{4cost dt}{\sqrt{16(1-sin^2 t)}}$[/tex] =[tex]$\int \frac{4cost dt}{\sqrt{16cos^2 t}}$[/tex]=[tex]$\int \frac{4cost dt}{4cost}$[/tex] =[tex]$\int dt$[/tex]

i don't see why I am getting a total wrong answer :S
 
  • #8
Slimsta said:
olympics distracted me for a while lol

yeah i got that antiderivative after googling it hhh.. my book didnt say anything about it.

now I am stuck on this question:
[tex]$\int \frac{dx}{\sqrt{16-x^2}}$[/tex]

i let x = 4sint
dx= 4cost dt

so
[tex]$\int \frac{dx}{\sqrt{16-16sin^2}}$[/tex] =[tex]$\int \frac{4cost dt}{\sqrt{16(1-sin^2 t)}}$[/tex] =[tex]$\int \frac{4cost dt}{\sqrt{16cos^2 t}}$[/tex]=[tex]$\int \frac{4cost dt}{4cost}$[/tex] =[tex]$\int dt$[/tex]

i don't see why I am getting a total wrong answer :S
= t + C

Now undo your substitution. You had x = 4sint ==> x/4 = sint ==> t = ?
 
  • #9
Mark44 said:
= t + C

Now undo your substitution. You had x = 4sint ==> x/4 = sint ==> t = ?

oh that's what it is. so the answer would be [tex]$\arcsin (x/4)+C$[/tex]

now my problems get more complicated, [tex]$\int \sqrt{5+4x-x^2}dx$ [/tex]
my textbook shows a simpler question and it works there but this question is a bit different.
i know that the first step is putting it like this: [tex]$\int \sqrt{5+(4x-x^2)}dx$ [/tex]
then something has to come out and then the substitution comes in..
whats my next step though?
 
  • #10
Write the part in the radical as 5 - (x^2 - 4x). Then complete the square so that you get something that looks like a - u^2 in the radical. Use a trig substitution on that.
 
  • #11
Mark44 said:
Write the part in the radical as 5 - (x^2 - 4x). Then complete the square so that you get something that looks like a - u^2 in the radical. Use a trig substitution on that.

i was thinking about this 5 - (x2 - 4x) and i was thinking, if i let y2 = x then
5 - (x2 - (2y)2)

would that work? because now I am on x2−a2
im not sure though
 
  • #12
I don't think that would work. Try what I suggested. If you had sqrt{3 + 2x - x^2)}, you would do this:
sqrt{3 + 2x - x^2)} = sqrt{3 - (x^2 - 2x)}
= sqrt{3 - (x^2 - 2x + 1) + 1} = sqrt{4 - (x - 1)^2}

If you let u = x - 1, you have sqrt{4 - u^2} and you're ready to do a trig substitution.

For my example, draw a right triangle with the hypotenuse labeled 2, the opposite side labeled u, and the adjacent side labeled sqrt{4 - u^2}. One acute angle would be labeled theta or t or whatever (but not x or u, since those are already in use).

Do the same on your problem.
 
  • #13
thats the exact same example from my textbook..
so what i tried is, 5 - (x^2 - 4x +1) + 1
= 6 - (x^2 - 4x +1)
how do you make it ^2?
 
  • #14
okay so i got
9 - (x-2)^2

now i substitute u=x-2, du=dx
i get [tex] $\int \sqrt{9 - u^2}du$ [/tex]

then u=3sint
du = 3cost dt
[tex] $\sqrt{5+4x-x^2}dx$[/tex] = 3cost
so [tex] $\int 3costdu$ [/tex]

am i right so far?
 

What is a trigonometric substitution?

A trigonometric substitution is a technique used in calculus to simplify integrals involving expressions with radicals or expressions with the form a2 - x2. It involves substituting a trigonometric function for the variable in the integral to make it easier to solve.

When should I use a trigonometric substitution?

A trigonometric substitution is most commonly used when the integral being solved involves an expression with a square root, or when there is a quadratic expression in the form a2 - x2 in the integrand. It can also be used to simplify integrals involving other types of expressions, such as x2 + a2 or x2 - a2.

What are some common trigonometric substitutions?

Some common trigonometric substitutions include x = a sin θ, x = a cos θ, and x = a tan θ, where a is a constant. The specific substitution used will depend on the form of the integral being solved and the trigonometric identities that can be applied.

How do I know if a trigonometric substitution is the right choice?

One way to determine if a trigonometric substitution is appropriate is to look at the form of the integrand. If it involves a square root, a quadratic expression, or other trigonometric functions, a trigonometric substitution may be helpful. It is also important to check if the integral can be solved using other techniques, such as integration by parts or u-substitution.

Are there any special considerations when using trigonometric substitutions?

Yes, there are a few things to keep in mind when using trigonometric substitutions. First, it is important to choose the appropriate trigonometric function to substitute in order to simplify the integral. Secondly, it is important to use the appropriate trigonometric identities to simplify the integral further. Lastly, it is important to pay attention to the limits of integration and make the necessary adjustments when substituting for the variable.

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