Expert: Paul Klarreich - 9/24/2007

Question
How would I integrate sin(sqrt x)dx?

Answer
Questioner:   Emily
Category:  Calculus
Private:  No
 
Subject:  Calculus - Integration by Parts
Question:  How would I integrate sin(sqrt x)dx?
.....................................
Hi, Emily,

For your integral,

{
| sin(sqrt(x)) dx
}

I'll outline what you must do; some of the work is hard to show in the rather klutzy interface we have here, so I'll leave that to you, such as taking care of the minus in between the 'parts', and maybe a 1/2 or a 2 factor here or there.

Preliminary -- keep in mind that the derivative of  sqrt(x) is 1/2sqrt(x).

----------------------------------------------------
First I will rewrite your integral:

{
| sin(sqrt(x)) dx =
}

multiplying top and bottom by  2 sqrt(x):

{ 2 sqrt(x) sin(sqrt(x)) dx
| -----------------------
}    2 sqrt(x)

Now the sin(sqrt(x)) can be the thing we INTEGRATE, the dv, instead of the u, and we can integrate that by simple substitution (what you normally call u-substitution, but we use 'u' for something else here).

u = sqrt(x),  and so   du = dx/2sqrt(x)

dv = sin(sqrt(x)) dx / 2sqrt(x) =   sin(z) dz,  and so
v = - cos z = - cos(sqrt(x))

uv = - sqrt(x) cos(sqrt(x))  << uv part of IBP

     - cos(sqrt(x)) dx
vdu = ---------------
       2 sqrt(x)

{
| vdu =
}

{  - cos(sqrt(x)) dx
| ---------------
}    2 sqrt(x)

{
| - cos z dz
}

= - sin(sqrt(x))

Now you can put those together to get your integral.