09/12/07 Lecture 7 Let’s take another look at the example that was mentioned in the last lecture summary:
ex cos xdx Two successive integrations by parts give us: ex cos xdx = ex (sin x) dx = x e sin x − (ex ) sin xdx ex sin x − ex (− cos x) dx = x x e sin x − e (− cos x) − (ex ) (− cos x)dx = ex sin x + ex cos x − ex cos xdx This implies that:
ex cos xdx = ex sin x + ex cos x − and the final answer is:
ex cos xdx
ex cos xdx =
1 x (e sin x + ex cos x) + C 2
A similar approach can be adopted in certain other cases when one does not even come up with a definite answer. Consider the integral:
cosn xdx where n is a positive integer number. We are not going to compute it directly. We will give an inductive formula instead, i.e. a formula that can be applied 1
a finite number of times for each specific problem of the above type (n will be given in concrete terms) and give the answer. To this end, write:
cosn xdx = cosn−1 x cos xdx = cosn−1 x(sin x) dx = cosn−1 x sin x − cosn−1 x sin x − (cosn−1 x) sin xdx = (n − 1) cosn−2 x(− sin x) sin xdx = cosn−2 x sin2 xdx = cosn−2 x(1 − cos2 x)dx = cosn−2 xdx − (n − 1) cosn xdx
cosn−1 x sin x + (n − 1) cosn−1 x sin x + (n − 1) cosn−1 x sin x + (n − 1)
So we have found that:
cosn xdx = cosn−1 x sin x + (n − 1)
cosn−2 xdx − (n − 1)
cosn xdx
and if we move all
cosn xdx’s to the left hand side and divide by n, we get: 1 n−1 cosn−1 x sin x + n n
cosn xdx =
cosn−2 xdx
These reduction formula can be used to find the indefinite integral in any concrete case. For example, if n = 4, one will have to apply this formula twice.
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