What's the deal with Integration by Parts?

EXPLANATION:

Integration by parts states that  ∫uv' dx = uv -  ∫u'v dx, where u & v are funcions of x and the notation u' means du/dx.

to do integration by parts given an integral  ∫f(x) dx, it involves writing f(x) as f(x) = u(x)v'(x), and then following the formula by determining u' and v.

The whole point of IBP is that  ∫u'v dx is hopefully easier to integrate than ∫uv' dx.

eg(1) take  ∫ xsin(x) dx. This is quite hard to integrate directly, so we use integration by parts. When choosing which is u(x) and v'(x), remember that you will have to integrate v' and differentiate u later. Often what happens is when you differentiate u, u'(x) turns out to be 1, which leaves you with a simple integration.

so take u(x) =x and v'(x)=sin(x)

=>u' =1,  v = -cos(x) (don't worry about the "+c", it's included at the end)

so following the formula: ∫xsin(x)dx = -xcos(x) - ∫-cos(x) = xcos(x) + ∫cos(x)dx = xcos(x) + sin(x) + c

this is the final answer to that particular question, and we see that integration by parts gives us another method of integtration

METHOD:

1) for ∫ f(x) dx, choose suitable functions u(x) and v'(x) such that f(x) = u(x) * v'(x). 

2) determine u'(x) and v(x) by differentiating and integrating respectively

3) use the formula ∫uv' dx = uv -  ∫u'v dx to find the answer!

DERIVATION: (not usually necessary for exam but interesting to see!)

differentiation by parts works like this, for, u & v as functions of x,

where u' = du/dx etc.

d/dx(uv) = (uv)' = uv' + u'v (proof omitted)

if we integrate both sides wrt x

=> uv =  ∫uv' dx +  ∫u'v dx

=>  ∫uv' dx = uv -  ∫u'v dx

:)