Expert: Paul Klarreich - 8/20/2009

Question
integrate the square root of 16-x^2 dx from -4 to 4 and x=4sin u

Answer
Questioner: Anna
Country: South Africa
Category: Calculus
Private: No
Subject: Integration by parts
Question: integrate the square root of 16-x^2 dx from -4 to 4 and x = 4 sin u
........................................
Why are you writing 'by parts'?  If you are doing  x = 4 sin u, (which does look like the right thing) this does not seem like IBP.

{
| sqrt(16-x^2) dx
}

x = 4 sin u
dx = 4 cos u du    << oops, I left out the du.
sqrt(...) = 4 cos u

Now your integral is:

{
| 16 cos^2(u) du
}

which you do by the 'half-angle trick'.

BTW, your integral appears to be just the upper semicircle of a circle with r=4, center at the origin, so the answer should be (1/2) 16pi = 8pi.
Check it out.