# [P3] Integration - by parts?Watch

This discussion is closed.
#1
I have an equation:

y = 2x^(1/2).Sin(x)

Need to integrate it. Not fully sure how to though. If i attempt to do it by parts, it seems to me like it will go around in circles.

Anyone care to shed some light on it? Thanks.

MdSalih
0
13 years ago
#2
that does seem rather peculiar.

In what context did u find this question? Are you sure you aren't meant to be doing integral of y² (volume of revolution) ?

if not, i don't know how you would do it.
0
#3
nup.. not a volume :-/ - It's just a straight integral...

MdSalih
0
13 years ago
#4
(Original post by MdSalih)
I have an equation:

y = 2x^(1/2).Sin(x)
Let u = 2x^(1/2) -> du/dx = (-1/2)(2)x^(-1/2) = -x^(-1/2)
Let dv/dx = sinx -> v = -cosx

Int. y dx. = -2x^(1/2).cosx - Int. x^(-1/2).cosx. dx.

To find Int. x^(-1/2).cosx dx.
Let u = cosx -> du/dx = -sinx
Let dv/dx = x^(-1/2) -> v = 2x^(1/2)

-> Int. x^(-1/2).cosx dx. = 2x^(1/2)cosx + 2 Int. x^(-1/2)sinx dx. + c

Hence: Int. y dx. = -2x^(1/2).cosx - 2x^(1/2)cosx - 2 Int. x^(-1/2)sinx dx. + d
-> 2 Int. x^(1/2).sinx dx. = -2x^(1/2).cosx - 2x^(1/2)cosx - 2 Int. x^(-1/2)sinx dx.
-> 4 Int. x^(1/2).sinx dx. = -4x^(1/2)cosx + d

-> Int. x^(1/2).sinx dx = k - x^(1/2).cosx
0
13 years ago
#5
(Original post by Nima)
Let u = 2x^(1/2) -> du/dx = (-1/2)(2)x^(-1/2) = -x^(-1/2)
Although I haven't read the rest of your solution (although it may well be right otherwise), when you differentiate 2x^0.5, im pretty sure u get x^(-0.5) and not -x^(-0.5).
0
13 years ago
#6
(Original post by Nima)
Let u = 2x^(1/2) -> du/dx = (-1/2)(2)x^(-1/2) = -x^(-1/2)
Let dv/dx = sinx -> v = -cosx

Int. y dx. = -2x^(1/2).cosx - Int. x^(-1/2).cosx. dx.

To find Int. x^(-1/2).cosx dx.
Let u = cosx -> du/dx = -sinx
Let dv/dx = x^(-1/2) -> v = 2x^(1/2)

-> Int. x^(-1/2).cosx dx. = 2x^(1/2)cosx + 2 Int. x^(-1/2)sinx dx. + c

Hence: Int. y dx. = -2x^(1/2).cosx - 2x^(1/2)cosx - 2 Int. x^(-1/2)sinx dx. + d
-> 2 Int. x^(1/2).sinx dx. = -2x^(1/2).cosx - 2x^(1/2)cosx - 2 Int. x^(-1/2)sinx dx.
-> 4 Int. x^(1/2).sinx dx. = -4x^(1/2)cosx + d

-> Int. x^(1/2).sinx dx = k - x^(1/2).cosx
As the above poster said, there was a slip when differentiating 2x^0.5
If you were to do it again with the error corrected you'd find that when you do your second integration by parts you're simply reversing the first one and your string of function on the right would cancel out, so you won't have anything on the right.
Can i ask where the question is from?
0
13 years ago
#7
yup right substitution, jus the diff. error otherwise all good.
0
13 years ago
#8
Web Mathematica baulks at it.

see

http://www.calc101.com/webMathematica/integrals.jsp

...which is usually ominous.

Aitch
0
13 years ago
#9
Int 2x^(1/2).Sin(x) dx
= 2 Int [x^(1/2)]sinx dx
==========================
Int [x^(1/2)]sinx dx

t = sqrtx => x = (t^2) => dx = 2t dt

Int [x^(1/2)]sinx dx
= Int t.sin(t^2).2t dt
= Int 2(t^2).sin(t^2) dt

ergo...

Int 2x^(1/2).Sin(x) dx
= 2 Int 2(t^2).sin(t^2) dt
= Int 4(t^2).sin(t^2) dt

u' = 4sin(t^2)
u = -(2/t])cos(t^2)
v = (t^2)
v' = 2t

Int (t^2).4sin(t^2) dt
= [(t^2)][-(2/t])cos(t^2)] - Int [2t][-(2/t])cos(t^2)]
= -(2t)cos(t^2) - Int [-4cos(t^2)]
= -(2t)cos(t^2) + Int 4cos(t^2)
= -(2t)cos(t^2) + (2/t)sin(t^2)
= (2/t)sin(t^2) - (2t)cos(t^2)

t = sqrtx, ergo:

(2/t)sin(t^2) - (2t)cos(t^2)) = 2(x^[-1/2])sinx - 2(x^[1/2])cosx

Int 2x^(1/2).Sin(x) dx = 2(x^[-1/2])sinx - 2(x^[1/2])cosx + C
0
13 years ago
#10
u' = 4sin(t^2)
u = -(2/t)cos(t^2)
There's a problem here. t isn't a constant so you can't do that.
If you were to differentiate 'u' you would have to use the product rule and would get something different to u'.
0
13 years ago
#11
(Original post by Gaz031)
There's a problem here. t isn't a constant so you can't do that.
If you were to differentiate 'u' you would have to use the product rule and would get something different to u'.
doh! - i just worked it out...
0
13 years ago
#12
Is it not just:

4/3x^3/2.sinx + C
0
13 years ago
#13
(Original post by !Laxy!)
Is it not just:

4/3x^3/2.sinx + C
Unfortunately not.
Differentiate it, using the product rule, and you'll see it's not the same as the original integral.
0
X
new posts
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### University open days

• University of East Anglia
All Departments Open 13:00-17:00. Find out more about our diverse range of subject areas and career progression in the Arts & Humanities, Social Sciences, Medicine & Health Sciences, and the Sciences. Postgraduate
Wed, 30 Jan '19
• Aston University
Wed, 30 Jan '19
• Solent University
Sat, 2 Feb '19

### Poll

Join the discussion

Remain (1019)
79.18%
Leave (268)
20.82%