Trigonometric Substitutions - help

[Slimsta]

Homework Statement

$\int \frac{\sqrt{1-4x^2}dx}{x}$

Homework Equations

The Attempt at a Solution

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

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

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

am i even on the right truck?

 

[Mark44]

When you do your trig substitution, replace every x in your original integral. x is still appearing in the denominator.

 

[Slimsta]

$\int \frac{cos\phi dx}{2sin\phi}$ =
$.5 \int \frac{cos\phi dx}{sin\phi}$ =
$.5 lnsin\phi + C$

but i don't think that's right is it..

 

[Mark44]

Nope. Start again from the beginning, and exchange x and dx for phi and dphi all in one step.

 

[Slimsta]

so i did the same thing over again and i got this:
$ (1/4) \int 1/sin\phi - sin\phi d\phi $
which =
$ (1/4)[ \int 1/sin\phi d\phi + cos\phi]$
right? now what..

 

[Mark44]

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
\int \left[\frac{1}{sin t} - sin t\right]dt
=\int \frac{1}{sin t}dt - \int sin t dt

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.

 

[Slimsta]

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:
$\int \frac{dx}{\sqrt{16-x^2}}$

i let x = 4sint
dx= 4cost dt

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

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

 

[Mark44]

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:
$\int \frac{dx}{\sqrt{16-x^2}}$

i let x = 4sint
dx= 4cost dt

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

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 = ?

 

[Slimsta]

= 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 $\arcsin (x/4)+C$

now my problems get more complicated, $\int \sqrt{5+4x-x^2}dx$
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: $\int \sqrt{5+(4x-x^2)}dx$
then something has to come out and then the substitution comes in..
whats my next step though?

 

[Mark44]

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.

 

[Slimsta]

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

 

[Mark44]

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.

 

[Slimsta]

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?

 

[Slimsta]

okay so i got
9 - (x-2)^2

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

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

am i right so far?