Question

Integration Q

Answer 1

\[\int\limits_{0}^{2} \sqrt{\frac{ x }{ 4-x }}dx \] \[(x=4\sin^2(\theta))\]

Answer 2

\[\int\limits_{0}^{2} \sqrt{\frac{ 4\sin^2(\theta) }{ 4 - 4\sin^2(\theta) }} dx\]

Answer 3

use:
1- sin^2 = cos^2

Answer 4

uhm where?

Answer 5

in the numerator (factor it first)

Answer 6

4 - 4sin^2 = 4(1-sin^2) . . .

Answer 7

* denominator

Answer 8

.

Can you simplify the fraction ?

Answer 9

\[\int\limits_{0}^{2}\sqrt{\frac{ 4(1-\cos^2(\theta)) }{ ? }}\] that is what you are suggesting me to do?

Answer 10

\[\int\limits_{0}^{2}\sqrt{\frac{ 4-4\cos^2(\theta) }{ 4-4\sin^2(\theta) }}\]

Answer 11

not quite
sorry, i mean this:

\[\frac{ 4\sin^2(\theta) }{ 4 - 4\sin^2(\theta) }=\frac{ 4\sin^2(\theta) }{ 4(1 - \sin^2(\theta)) } = ...\]

Answer 12

ohh alright haha

Answer 13

(i got my words mixed up numerator/denominator )

Answer 14

in that case I can cancel both 4's, correct.

Answer 15

which I will be left with sin^2(theta) / 1-sin^2(theta) <-- cost^2(theta)

Answer 16

which then becomes tan^2(theta)

Answer 17

Then I have to solve tan^2(2) - tan^2(0) right? or.. first I have to get back what was x

Answer 18

dont forget the sqrt

Answer 19

and you will also want to change dx to d(theta)

Answer 20

\[\int\limits\limits_{0}^{2} \sqrt{\tan^2(\theta)}dx\]

Answer 21

good so what is \(dx/d\theta\) ?

Answer 22

remember

\[x = 4\sin^2\theta\]

Answer 23

that means we have to solve for theta? that was we bring back what theta is

Answer 24

x*

Answer 25

don't solve it yet, differentiate first

Answer 26

\[dx = \quad . . .\quad d\theta\]

Answer 27

8sin(theta) cos(theta)?

Answer 28

yes

Answer 29

so, now swap dx for 8sin(theta) cos(theta) d(theta) in the integral

Answer 30

ooo I see what you did there.

Answer 31

\[\int\limits_{0}^{2}\sqrt{\tan^2(\theta)} *8\sin(\theta)\cos(\theta) d \theta\]

Answer 32

right.

can we simplify the integrand ?

Answer 33

\[\int\limits_{0}^{2}8\sin(\theta)\cos(\theta)\sqrt{\tan^2(\theta)}d \theta\]\[8\int\limits_{0}^{2} \sin(\theta)\cos(\theta)\sqrt{\tan^2(\theta)} d \theta\]

Answer 34

a good start,

now also recall that \(\sqrt a^2 = a\)

and that tan = sin/cos

Answer 35

\[8\int\limits_{0}^{2}\sin(\theta)\cos(\theta)\tan(\theta) d \theta\]

Answer 36

Forgot to simplify that hehe

Answer 37

\[\sin(\theta)\cos(\theta)\tan(\theta)\\ = \sin(\theta)\cos(\theta)\frac{\sin(\theta)}{\cos(\theta)}\\= . . .\]

Answer 38

\[8\int\limits_{0}^{2}\sin(\theta)\cos(\theta)\frac{ \sin(\theta) }{ \cos(\theta) } d \theta\]

Answer 39

we kill cos(theta) and we are left with sin^2(theta)

Answer 40

\[8\int\limits_{0}^{2} \sin^2(\theta) d \theta\]

Answer 41

great work,

(also. we might want to explicitly write it as:
\[8\int\limits_{x=0}^{x=2} \sin^2(\theta) d \theta\])

so we don't forget where those limits really are

Answer 42

alright

Answer 43

can you integrate
\[\int \sin^2(\theta)\, d \theta\]?

Answer 44

I'm not sure how to integrate sin^2(theta)

Answer 45

according to the table of integrals that its equals to x/2 - sin(2ax)/4a

Answer 46

\[8[\frac{ \theta }{ 2 } - \frac{ \sin(2\theta) }{ 4 }]\]

Answer 47

and remembering our limits: \[8\left[\frac{ \theta }{ 2 } - \frac{ \sin(2\theta) }{ 4 }\right]_{x=0}^{x=2}\]

Answer 48

correct. Now what comes in mind is substitute the thetas with x's

Answer 49

or changing the limits into theta form

Answer 50

either should work, one way is usually easier

Answer 51

changing the x values to the theta values

Answer 52

x = 4sin^2(theta)

Answer 53

so x=0 ---> theta = . . .
and
x=2 ----> theta = . . .

Answer 54

I have to solve for theta first.. which is basically the same as if I substituted it uhmm

Answer 55

x/4 = sin^2(theta)

Answer 56

0 = sin^2(theta)
1/2 = sin^2(theta)

Answer 57

when sin is 0 that is pi?

Answer 58

no. that is 0

Answer 59

How do you figure out when sin is squared?