Integrals using "by parts approach"

The general formula:

INT(u*dv) = u*v - INT(v*du)

You have to find:
u =
du =
v =
dv =

You do this by choosing dv to be the most complicated part you can integrate. Then, u is set to the rest. I'll choose cos(x) for dv (and dv gets the dx).

u = x^2
du =
v =
dv = cos(x)*dx

We haven't changed anything:

INT(x^2cos(x)*dx)=INT(u*dv)

Now we fill in the rest:

u = x^2
du = 2x*dx
v = sin(x)
dv = cos(x)*dx

And write out the fomula:

INT(u*dv) = u*v - INT(v*du)
INT(u*dv) = x^2sin(x) - INT(sin(x)*2x*dx)

So what has happened is we can't solve it yet. You'll often find that integration by parts is a multi-pass algorithm.

Now we focus on INT(sin(x)*2x*dx), use same procedure:

u = 2x
du = 2*dx
v = -cos(x)
dv = sin(x)*dx

INT(u*dv) = u*v - INT(v*du)
INT(u*dv) = 2x*-cos(x) - INT(-cos(x)*2*dx)

Now this part is solvable.

INT(-cos(x)*2*dx) = -2*INT(cos(x)*dx) = -2sin(x)

So replacing that into most recent INT(u*dv) formula:

INT(u*dv) = 2x*-cos(x) - -2sin(x) = 2sin(x) - 2xcos(x)

Now we replace that back into the original (this is the original u and v now):

INT(u*dv) = x^2sin(x) - (2sin(x) - 2xcos(x))

**oh by the way I've ignored all my +C's since I know I'll be evaluating it at the end**

Now I can evaluate it (the integral has disappeared, just left with x's now):

x^2sin(x) - (2sin(x) - 2xcos(x)) | [0,pi/2]

I get 0.4674 once you plunk and crank it (the zero has no effect because it just eliminates each term).

Remember to put your calculator in radians.

**Note: this is the long way that works for all cases. There's acutally a slick table method you can do too if one of the parts is a polynomial (like in this one, x^2).**