Question

anyone can help me to evaluate this integrals?

Answer 1

what is the question?

Answer 2

\[\int\limits_{}^{}\frac{ ydy }{ \sqrt{a^2-y^2} }\]

Answer 3

You have learned «u-substitution», correct?

Answer 4

is it necessary to use u-substitution because my instructor doesnt teach it yet..

Answer 5

There is an alternative way that is usually taught before «u-substitution», and that is «recognizing derivatives»...

Answer 6

Is that what you do? ... try to recognize a derivative(?)
- i.e. what function would give the derivative that would be this integrand?

Answer 7

i think i can use the power formula here?

Answer 8

Power rule? No, you can't.

Answer 9

i will set the denominator as (a^2-y^2) ^-1/2?

Answer 10

Integrals don't quite work that way. Through u-substitution, there is a way to make the integrand just, \(\color{blue}{u^{-1/2}}\).

But, you haven't learned u-substitution, as you said, so we need to 'recognize the derivative'.

Answer 11

well, there would be a negative 1/2 there, actually, but in any case, we can't apply that yet, since you don't yet know the method.

Answer 12

Tell me this.

What is the derivative of \(\sqrt{~x~}\) with respect to x?

Answer 13

1/2x^-1/2 ?

Answer 14

Yes,

\(\color{#000000 }{ \displaystyle \frac{d}{dx}~\sqrt{x~}=\frac{1}{2\sqrt{x}} }\)

Answer 15

And then, by the chain rule principle, what is,

\(\color{#000000 }{ \displaystyle \frac{d}{dx}~\sqrt{f(x) }=? }\)

Answer 16

f(x) means?

Answer 17

f(x) is just some [differentiable] function of x.

Answer 18

so i think its the same at what i give,but instead of x , put f(x)?

Answer 19

Not exactly.

Answer 20

What is the derivative of

ln( x^2 +1) ?

Answer 21

1/(x^2+1)

Answer 22

Oh I see what we are missing

Answer 23

2x

Answer 24

yes, and where?

Answer 25

numerator

Answer 26

yes, very good, and that is the chain rule.

Answer 27

So, back to what I asked,

\(\color{#000000 }{ \displaystyle \frac{d}{dx}~\sqrt{f(x)}=\frac{f'(x)}{2\sqrt{f(x)}} }\)

wouldn't you agree?

Answer 28

\(f'(x)\) is a notation for the [first] derivative of a function f(x).

Answer 29

yup,im taking down

Answer 30

So, will just do a clarification, that you know if,

\(\color{#000000 }{ \displaystyle \frac{d}{dx}~\sqrt{f(x)}=\frac{f'(x)}{2\sqrt{f(x)}} }\)

then,

\(\color{#000000 }{ \displaystyle \int \frac{f'(x)}{2\sqrt{f(x)}}~dx=\sqrt{f(x)}~\color{grey}{\rm +C} }\)

Am I correct?

Answer 31

hmmm... how about the y on the numerator on the given problem?

Answer 32

We will explore that. CAn you confirm that you understand the above post?

Answer 33

yup, thats correct

Answer 34

You want to find the following integral

\(\color{#000000 }{ \displaystyle\int \frac{y}{\sqrt{a^2-y^2 }}~dy }\)

by recognizing the derivatives,

Answer 35

So, to get that denominator, what would you put inside the square root?

\(\color{#000000 }{ \displaystyle \frac{d}{dx}~\sqrt{~??~}=\frac{?}{2\sqrt{~??~}}}\)

Answer 36

a^2-y^2

Answer 37

yes, very good

Answer 38

-2y for numerator

Answer 39

\(\color{#000000 }{ \displaystyle \frac{ d }{dy} ~\sqrt{a^2-y^2}~=~? }\)

Answer 40

-2y/ 2sqrt of a^2-y^2

Answer 41

and simplify that please

Answer 42

2/2 =1 :)

Answer 43

\[\frac{ -y }{ \sqrt{a^2-y^2} }\]

Answer 44

yes, very good. So,

\(\color{#000000 }{ \displaystyle \frac{ d }{dx} ~\left[\color{blue}{\sqrt{a^2-y^2}}~\right]~=~\frac{-y}{\sqrt{a^2-y^2}} }\)

Answer 45

Your integrand doesn't have a negative, and that is an easy problem to take care off.

Answer 46

What function would give you?

\(\color{#000000 }{ \displaystyle \frac{ d }{dx} ~\left[\color{blue}{~?}~\right]~=~\frac{y}{\sqrt{a^2-y^2}} }\)

(without the negative)

Answer 47

sqrt of a^2-y^2

Answer 48

that, when differentiated, will have a negative (a minus) on top.
Right?

Answer 49

\(\color{#000000 }{ \displaystyle \frac{ d }{dx} ~\left[\color{blue}{\sqrt{a^2-y^2}}~\right]~=~\frac{-y}{\sqrt{a^2-y^2}} }\)

Answer 50

yeah

Answer 51

\[-\sqrt{a^2-y^2}\]

Answer 52

Yes, very good.

Answer 53

And, since this is an integral you add +C to the result

Answer 54

yeah, i see, thats the answer right?

Answer 55

yes

Answer 56

thanks solomon,i have learned alot

Answer 57

So, since

\(\color{#000000 }{ \displaystyle \frac{ d }{dx} ~\left[\color{blue}{-\sqrt{a^2-y^2}}~\right]~=~\frac{y}{\sqrt{a^2-y^2}} }\)

therefore

\(\color{#000000 }{ \displaystyle \int \frac{y}{\sqrt{a^2-y^2}}dy =\color{blue}{-\sqrt{a^2-y^2}}\color{grey}{~\rm +C} }\)

Answer 58

yw

Answer 59

i can do it with u-substitution but my instructor never tough it yet ^^

Answer 60

Yes, very conveniently....

Answer 61

I will use x instead of y, i like it that ay better.

Answer 62

can you help me on another problem ?^^

Answer 63

Yes, I can try

Answer 64

thanks

Answer 65

yw