small question regarding integrals

Original post by Kolasinac138

Through the integration process I am going over on a worksheet, in the integral of

\dfrac {cos(x)}{\sqrt{(2+sinx)}}

I'm told that cosx is not part of the integrand because it is the derivative of 2+sinx, may I ask why this would be the case?

There's a rule:

\int \frac{f'(x)}{f(x)} dx= ln|f(x)| + c

Do you think you'll be able to do it now - or rather get why this would be the case?

(edited 9 years ago)

Reply 2

OP

Original post by kingaaran

There's a rule:

\int \frac{f'(x)}{f(x)} dx= ln|f(x)| + c

Do you think you'll be able to do it now - or rather get why this would be the case?

Well the derivative of ln(x) is 1/x or 2/2x or etc etc, but I don't understand how that helps here.

Reply 3

crashMATHS

Original post by Kolasinac138

Well the derivative of ln(x) is 1/x or 2/2x or etc etc, but I don't understand how that helps here.

The way I've been taught is that if you have a function, say

\frac{2x}{x^2}

and you integrated it with respect to x, you can use that formula, since the derivative of the denominator is equal to the numerator. But also, if let's say we had

\frac{4x}{x^2}

and integrated that, I could still use that principle, but all I would do is have two lots of

2ln |x^2| +c

, because the numerator is two lots of the derivative of the denominator. Does that help?

Posted from TSR Mobile

Reply 4

OP

Original post by kingaaran

The way I've been taught is that if you have a function, say

\frac{2x}{x^2}

and you integrated it with respect to x, you can use that formula, since the derivative of the denominator is equal to the numerator. But also, if let's say we had

\frac{4x}{x^2}

and integrated that, I could still use that principle, but all I would do is have two lots of

2ln |x^2| +c

, because the numerator is two lots of the derivative of the denominator. Does that help?

Posted from TSR Mobile

Well, I mean, the answer here is 2

\sqrt{2+sin(x)}

or something similar and the main point is that cos(x) isn't in the integrand but still not sure why.

I mean, I understand integration by substitution but this is meant to be done without that knowledge.

Reply 5

Shahriyar

1

u=sin (x)

\frac{du}{dx}=cos(x)

dx=

\frac {du}{cos(x)}

\int \frac{cos(x)}{\sqrt(1+sin(x))}

\frac {du}{cos(x)}

so cos (x) cancels out

\int \frac{1}{\sqrt(1+u)}

this is sort of the proof of the rule that kingaaran said :smile:

when you have derivative of denominator in numerator substitution always works like this
though in this case needs a little bit to be played around with

(edited 9 years ago)

Reply 6

crashMATHS

Original post by Kolasinac138

Well, I mean, the answer here is 2

\sqrt{2+sin(x)}

or something similar and the main point is that cos(x) isn't in the integrand but still not sure why.

I mean, I understand integration by substitution but this is meant to be done without that knowledge.

Oh, I overlooked the square root sign in your first post - I was just referring to if you had cos(x)/(2+sin(x). With this one, it might have something to do with the cos(x) cancelling out when you use the integration by substation formula, but if not, I'm not really too sure :/

Posted from TSR Mobile

(edited 9 years ago)

Reply 7

OP

Original post by kingaaran

Oh, I overlooked the square root sign in your first post - I was just referring to if you had cos(x)/(2+sin(x). With this one, it might have something to do with the cos(x) cancelling out when you use the integration by substation formula, but if not, I'm not really too sure :/

Posted from TSR Mobile

Well, I'm not too sure either.

Stating that cos(x) is the derivative of 2+sin(x) therefore meaning cos(x) is excluded from the integrand doesn't make a whole load of sense to me.

A derivative multiplied by the integral of that derivative is ... :biggrin:

Reply 8

atsruser

11

Original post by Kolasinac138

Through the integration process I am going over on a worksheet, in the integral of

\dfrac {cos(x)}{\sqrt{(2+sinx)}}

I'm told that cosx is not part of the integrand because it is the derivative of 2+sinx, may I ask why this would be the case?

What you have written above doesn't make much sense as the

\cos x

certainly is part of the integrand, since its value depends on the variable of integration. (it's not a constant that can be taken outside).

However, the observation essentially indicates the correct substitution to make:

u = 2+\sin x

.

Reply 9

Original post by Kolasinac138

Through the integration process I am going over on a worksheet, in the integral of

\dfrac {cos(x)}{\sqrt{(2+sinx)}}

I'm told that cosx is not part of the integrand because it is the derivative of 2+sinx, may I ask why this would be the case?

You do not seem to understand what Integrand means - it is the function to be integrated - so Cos(x) is certainly part of the integrand

This is an inverse chain rule question

Consider what happens when you differentiate

\sqrt{2+\sin x}

(edited 9 years ago)

Reply 10

atsruser

11

Original post by TenOfThem

You do not seem to understand what Integrand means - it is the function to be integrated - so Cos(x) is certainly part of the integrand

As has been explained this question uses

\displaystyle \int \dfrac{f'(x)}{f(x)} dx = ln|f(x)| + c

In your case - if you differentiate the denominator of your integrand you get a multiple of the numerators you have an integral of this format

But the integrand isn't of this form, due to the square root.

Reply 11

Original post by atsruser

But the integrand isn't of this form, due to the square root.

corrected

Reply 12

davros

Original post by Kolasinac138

Through the integration process I am going over on a worksheet, in the integral of

\dfrac {cos(x)}{\sqrt{(2+sinx)}}

I'm told that cosx is not part of the integrand because it is the derivative of 2+sinx, may I ask why this would be the case?

Either you, or the person who's advising you, seems to be confused about terminology.

As TenOfThem has pointed out, integrand just means "the thing what you are integrating" so it definitely includes the cos x.

There is an obvious substitution that will sort out this integral for you (and once you've done a few similar ones you can probably spot the answer by recognition).

Reply 13

Desk-Lamp

1

This is the fype of integral you can do easily by inspection once you spot it-
just consider the derivative of

2\sqrt{f(x)}

Reply 14

OP

Original post by atsruser

What you have written above doesn't make much sense as the

\cos x

certainly is part of the integrand, since its value depends on the variable of integration. (it's not a constant that can be taken outside).

However, the observation essentially indicates the correct substitution to make:

u = 2+\sin x

.

That is what I thought, personally, so was a little confused.

I suppose it's of the form

g'(x)f(g(x))

Although I'm supposed to be able to do this without knowledge of integration by substitution.

Given the derivative of

\sqrt{2+sin(x)}

would be

\dfrac{cosx}{2\sqrt{2+sinx}}

but I still don't see how it's relevant.

(edited 9 years ago)

Reply 15

Original post by Kolasinac138

Given the derivative of

\sqrt{2+sin(x)}

would be

\dfrac{cosx}{2\sqrt{2+sinx}}

but I still don't see how it's relevant.

You cannot see that it looks JUST LIKE your question apart from a factor of 2?

Reply 16

OP

Original post by TenOfThem

You cannot see that it looks JUST LIKE your question apart from a factor of 2?

No, I do understand that :biggrin:

and therefore the integral is

2\sqrt{2+sinx}

+c but still, just from basics I didn't understand why cosX was excluded.

Reply 17

Original post by Kolasinac138

I didn't understand why cosX was excluded.

excluded from what?

Reply 18

OP