Question

integrals,

can you please help me integrate sqrt(1 - 4x^2)

Answer 1

can i help

Answer 2

try the substitution n = 1 - 4x^2

Answer 3

what are the choices

Answer 4

I need to trigonometricly substitute

Answer 5

* u = 1 - 4x^2

Answer 6

to learn that kind of technique because I rly strucly with trigonometric substitutions

Answer 7

strugle *

Answer 8

can we please solve it with trigonometric substituion

Answer 9

so that I can get the concept

Answer 10

what are the choices so i can be sure thats 1 of them or do u have no choices

Answer 11

yea - i'm trying to remember the appropriate substiuttion here

Answer 12

trig substitution is the choice @youn

Answer 13

this is not a multiple choice question, I just have to solve it @youn

Answer 14

i think its x = sin theta

Answer 15

thats what i got

Answer 16

i did that but I cant get it to work @welshfella

Answer 17

\[\large\rm \int\limits \sqrt{1-4x^2}~dx=\int \sqrt{1-(2x)^2}~dx\]

Under the root your expression has the form: \(\large\rm 1-(stuff)^2\)
Notice that if you replace the "stuff" with sin(theta),
you can apply your Pythagorean Identity,\[\large\rm 1-(\sin \theta)^2=\cos^2\theta\]

Answer 18

or myhbe in this case its x = sin 2theta as we have 4x^2

Answer 19

ye but that stuff is 2x

Answer 20

2x = sin(theta)

Answer 21

Good good good :)\[\large\rm 2x=\sin \theta\]

Answer 22

hm

Answer 23

sin theta = 2x is the way to go

Answer 24

so then its sqrt(cos^2(theta) + 2cos(theta) ?

Answer 25

thats gives you sqrt ( 1 - sin^2 theta)

Answer 26

\[\rm \int\limits\sqrt{1-(\color{orangered}{2x})^2}~dx\quad=\quad \int\limits\sqrt{1-(\color{orangered}{\sin \theta})^2}~dx\quad=\quad \int\limits\sqrt{\cos^2\theta}~dx\]

Answer 27

or rather 2 * sqrt(cos(theta)) + 2cos(theta)

Answer 28

Still have to deal with the `dx` from that point,
but at least it makes the root easier, ya?

Answer 29

so use the trig identity relating sin^2 and cos^2

Answer 30

i dont get what I should do with dx

Answer 31

2x = sin(theta)

Answer 32

x = sin(theta)/2

Answer 33

dx = cos(theta/2 right ???

Answer 34

)*

Answer 35

yea - need to find dx is terms of a trig function

Answer 36

\[\large\rm dx=\frac{1}{2}\cos \theta~d \theta\]You moved the 2 to the other side? Good good good.

Answer 37

yes and I got 2 * sqrt(cos(theta)) + 2cos(theta)

Answer 38

is it good so far ????

Answer 39

Why do you have addition?

Answer 40

hmm

Answer 41

\[\rm \int\limits\limits\sqrt{1-(\color{orangered}{2x})^2}~dx\quad=\quad \int\limits\limits\sqrt{1-(\color{orangered}{\sin \theta})^2}~dx\quad=\quad \int\limits\limits\sqrt{\cos^2\theta}~dx\]From that point,\[\large\rm \int\limits \cos \theta~\color{orangered}{dx}\quad=\quad \int\limits \cos \theta~\color{orangered}{\frac{1}{2}\cos \theta~d \theta}\]

Answer 42

It should be a 1/2, not a 2, ya?

Answer 43

well dx = cos(x)/2

Answer 44

cos(theta)/2 dtheta

Answer 45

Yes,
you're replacing dx with... cos(theta)/2 dtheta

Answer 46

ok so 1/2 * cos^2(x) and then half angle formula and then done

Answer 47

Ummm, half angle,
and a little bit of work from there.
Undoing your substitution will require a triangle :) remember that step?

Answer 48

triangle or algebra

Answer 49

i think algebra works too

Answer 50

sure sure sure

Answer 51

2x = sin theta
x = (1/2) sin theta
dx = (1/2) cos theta

Answer 52

thanks

Answer 53

hi