# Trigonometric Substitutions - help!

1. Feb 15, 2010

### Slimsta

1. The problem statement, all variables and given/known data
$$\int \frac{\sqrt{1-4x^2}dx}{x}$$

2. Relevant equations

3. The attempt at a solution
im stuck and i have no idea why im 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?

2. Feb 15, 2010

### Staff: Mentor

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

3. Feb 15, 2010

### Slimsta

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

but i dont think that's right is it..

4. Feb 15, 2010

### Staff: Mentor

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

5. Feb 15, 2010

### 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..

6. Feb 15, 2010

### Staff: Mentor

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.

7. Feb 23, 2010

### 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 im 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 dont see why im getting a total wrong answer :S

8. Feb 24, 2010

### Staff: Mentor

= t + C

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

9. Feb 24, 2010

### Slimsta

oh thats 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?

10. Feb 24, 2010

### Staff: Mentor

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. Feb 24, 2010

### Slimsta

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 im on x2−a2
im not sure though

12. Feb 25, 2010

### Staff: Mentor

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. Feb 25, 2010

### 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?

14. Feb 27, 2010

### 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?