Integral by parts Q #6

Ahmed.Azuz

Could some one help me please :confused:

Integral by parts

X sin X

Reply 1

j.alexanderh

14

Which of those functions has the nicest derivative?

Reply 2

Ahmed.Azuz

OP

Sin X

Reply 3

j.alexanderh

14

Did you mean

\int xsinx\ dx

? The derivative of x is 1, which is very nice indeed.

Reply 4

dknt

12

Original post by Ahmed.Azuz

Sin X

Really? Think about it. If you set it out such that you end up integrating x and differentiating sinx you'll end up increasing the power of x and making it more confusing.

Reply 5

Ahmed.Azuz

OP

yeah I did, but isn't dy/dx Sin X = -cos x ?

Reply 6

dknt

12

Original post by Ahmed.Azuz

yeah I did, but isn't dy/dx Sin X = -cos x ?

the derivative of sinx is cosx, not -cosx, but even so why would you want to do that, when differentiating the x term would give you 1. With your sinx, you could choose to differentiate or integrate, but either way you'll end up cycling though cos and sin. But if you integrate your x term you'll increase the powers of x making it more confusing, so why not have your u=x and dv/dx=sinx.

(edited 13 years ago)

Reply 7

Ahmed.Azuz

OP

int x sin x dx = x(-cos x) - int (-cosx)dx= -x cos x + int cos x dx = -x cos x + sin x +c

isn't this correct? ''int = the integral sign''

Reply 8

mir3a

13

Original post by Ahmed.Azuz

yeah I did, but isn't dy/dx Sin X = -cos x ?

That is a lot harder working with, than if you were to find the derivative of X.
Let u=x
let dv/dx=sinx

Using the formula:

U*V- the intergral of (V*du/dx), is the correct method.

Reply 9

mjtriggs

10

Original post by Ahmed.Azuz

yeah I did, but isn't dy/dx Sin X = -cos x ?

dy/dx (sinx) = cosx

For Int. by parts, you need to choose one function as u, and one as v'.

I would use u=x and v'=sinx

Then u'=1, and v=-cosx.

Reply 10

j.alexanderh

14

Original post by Ahmed.Azuz

yeah I did, but isn't dy/dx Sin X = -cos x ?

Yes (well, it should be d/dx not dy/dx), but

\frac{d}{dx}x=1

. If you use this as the part you want to take the derivative of, there will be no x term in your integral and therefore you will have only a single trig function to integrate, which should be easy.

(edited 13 years ago)

Reply 11

Ahmed.Azuz

OP

thanks guys now I understand how to choose the nicest way, can u please check my comment #7 whether it's true or not

Reply 12

Pigeon93

2

let u=x and dv=sinx, so du=dx and v=-cosx

then since Int(uv) = uv - Int(v du)

you get -xcosx - Int(-cosx dx)
=> -xcosx + Int(cosx dx)
=> -xcosx + sinx + c

^^I hope that's right, somebody pls correct me if not. Haven't done this in ages.

If ever, i was taught the rule "ILATE" when it comes to deciding which function to let equal "u" (and not dv) to make integration by parts the easiest. "ILATE" stands for "Inverse, Logarithmic, Algebraic (polynomials), Trigonometric, Exponential", and you let "u" equal whichever of the two, multiplied functions being integrated comes first in that list. ex: x=A, whereas sinx=T, so u=x, and then let dv=sinx

Hope this helps!

Reply 13

mir3a

13

Original post by Pigeon93

let u=x and dv=sinx, so du=dx and v=-cosx

then since Int(uv) = uv - Int(v du)

you get -xcosx - Int(-cosx dx)
=> -xcosx + Int(cosx dx)
=> -xcosx + sinx + c

^^I hope that's right, somebody pls correct me if not. Haven't done this in ages.

If ever, i was taught the rule "ILATE" when it comes to deciding which function to let equal "u" (and not dv) to make integration by parts the easiest. "ILATE" stands for "Inverse, Logarithmic, Algebraic (polynomials), Trigonometric, Exponential", and you let "u" equal whichever of the two, multiplied functions being integrated comes first in that list. ex: x=A, whereas sinx=T, so u=x, and then let dv=sinx

Hope this helps!

Yep thats correct :smile: