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

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?

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

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 Generalisation Integrating: Simplifying the l.h.s.: Rearranging:

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

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

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

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 1st term, uv, is completed but the 2nd term still needs to be integrated. and differentiate integrate ( +C comes later )

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

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

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

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

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.

Using Integration by Parts Integration by parts cannot be used for every product. It works if we can integrate one factor of the product, 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.

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 , we will need to differentiate and integrate x. Tip: Whenever appears in an integration by parts we choose to let it equal u.

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

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

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

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.

e.g. 5 Find Solution: We write this as:

e.g. 5 Find So, where and We next use integration by parts for I2

e.g. 5 Find So, where and We next use integration by parts for I2

2 equations, 2 unknowns ( I1 and I2 ) ! e.g. 5 Find So, . . . . . ( 1 ) . . . . . ( 2 ) 2 equations, 2 unknowns ( I1 and I2 ) ! Substituting for I2 in ( 1 )

2 equations, 2 unknowns ( I1 and I2 ) ! e.g. 5 Find So, . . . . . ( 1 ) . . . . . ( 2 ) 2 equations, 2 unknowns ( I1 and I2 ) ! Substituting for I2 in ( 1 )

2 equations, 2 unknowns ( I1 and I2 ) ! e.g. 5 Find So, . . . . . ( 1 ) . . . . . ( 2 ) 2 equations, 2 unknowns ( I1 and I2 ) ! Substituting for I2 in ( 1 )

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

Solutions: and Let 1. . . . . . ( 1 ) and Let For I2: Subs. in ( 1 )

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