Integration by Parts We develop the formula by

Integration by Parts We develop the formula by considering how to differentiate products. If where u and v are both functions of x. Substituting for y, e. g. If ,

Integration by Parts So, Integrating this equation, we get The l. h. s. is just the integral of a derivative, so, since integration is the reverse of differentiation, we get Can you see what this has to do with integrating a product?

Integration by Parts Here’s the product. . . if we rearrange, we get The function in the integral on the l. h. s. . . . is a product, but the on the r. h. s. . is a simple function that we can integrate easily.

Integration by Parts Here’s the product. . . if we rearrange, we get So, we’ve integrated ! We need to turn this method ( called integration by parts ) into a formula.

Example Integration by Parts Integrating: Simplifying the l. h. s. : Rearranging: Generalisation

Integration by Parts SUMMARY Integration by Parts To integrate some products we can use the formula

Integration by Parts So, Using this formula means that we differentiate one factor, u to get . . .

Integration by Parts So, Using this formula means that we differentiate one factor, u to get . . . and integrate the other , to get v

Integration by Parts So, Using this formula means that we differentiate one factor, u to get . . . and integrate the other , to get v e. g. 1 Find Having substituted in the formula, notice that the andbut the 2 nd term still 1 st term, uv, is completed needs to be integrated. differentiate integrate ( +C comes later )

Integration by Parts So, and differentiate We can now substitute into the formula integrate

Integration by Parts So, and differentiate integrate We can now substitute into the formula The 2 nd term needs integrating

Integration by Parts e. g. 2 Find Solution: differentiate So, and integrate This is a compound function, so we must be careful.

Integration by Parts Exercises Find 1. Solutions: 1. 2. ò xe x dx =

Integration by Parts Definite Integration by Parts With a definite integral it’s often easier to do the indefinite integral and insert the limits at the end. We’ll use the question in the exercise you have just done to illustrate.

Integration by Parts Using Integration by Parts Integration by parts cannot be used for every product. It works if Ø we can integrate one factor of the product, Ø the integral on the r. h. s. is easier* than the one we started with. * There is an exception but you need to learn the general rule.

Integration by Parts e. g. 3 Find Solution: What’s a possible problem? ANS: We can’t integrate . Can you see what to do? If we let and differentiate and integrate x. Tip: Whenever , we will need to appears in an integration by parts we choose to let it equal u.

Integration by Parts e. g. 3 Find So, differentiate integrate The r. h. s. integral still seems to be a product! BUT. . . x cancels. So,

Integration by Parts e. g. 4 Solution: Let and The integral on the r. h. s. is still a product but using the method again will give us a simple function. We write

Integration by Parts e. g. 4 Solution: . . . (1) Let and So, Substitute in ( 1 )

Integration by Parts Example e. g. 5 Find Solution: It doesn’t look as though integration by parts will help since neither function in the product gets easier when we differentiate it. However, there’s something special about the 2 functions that means the method does work.

Integration by Parts e. g. 5 Find Solution: We write this as:

Integration by Parts e. g. 5 Find So, where and We next use integration by parts for I 2

Integration by Parts e. g. 5 Find So, . . . (1). . . (2) 2 equations, 2 unknowns ( I 1 and I 2 ) ! Substituting for I 2 in ( 1 )

Integration by Parts Exercises 1. 2. ( Hint: Although 2. is not a product it can be turned into one by writing the function as. )

Integration by Parts Solutions: Let 1. and . . . (1) For I 2: Let Subs. in ( 1 ) and

Integration by Parts 2. Let So, and This is an important application of integration by parts

Integration by Parts