|
|
|
Revision:Integration by PartsTSR Wiki > Study Help > Subjects and Revision > Revision Notes > Mathematics > Integration by Parts From the product rule, we can obtain the following formula, which is very useful in integration:
It is used when integrating the product of two functions (
Deriving the FormulaThe integration by parts formula is derived from the product rule, used in differentiation. If
Integrating each side with respect to x:
Rearranging:
Integrating ln xIntegration by parts provides us with the method for integrating
Let
Let
Putting it into the formula:
Uses in A-Level ExamsThe technique of integration by parts will often by indicated by explicitly stating so in the question. However, there is a chance you will have to decide for yourself, in which case it is useful to have an idea of when to use the method. An example of when to use integration by parts is when we have a x term, multiplied by a trigonometric or exponential function. Here is an example.
Let
Let
Putting it into the formula:
Comments |
|
||||||||||||||||
![\displaystyle \int^b_a u\frac{dv}{dx}dx=[uv]^b_a - \int^b_a v\frac{du}{dx}dx](http://www.thestudentroom.co.uk/latexrender/pictures/26/26f57e7030faafa94f93a25c41cc5dcb.png "\displaystyle \int^b_a u\frac{dv}{dx}dx=[uv]^b_a - \int^b_a v\frac{du}{dx}dx")
![I=[x(-\cos x)]_0^\pi - \displaystyle \int_0^\pi (-\cos x)1 \ dx](http://www.thestudentroom.co.uk/latexrender/pictures/45/4521d6ce03319f630a32ffc1d84b65ba.png "I=[x(-\cos x)]_0^\pi - \displaystyle \int_0^\pi (-\cos x)1 \ dx")
![I=\pi+[\sin x]_0^\pi](http://www.thestudentroom.co.uk/latexrender/pictures/e0/e02544f3ebe797e3e7f937c06184f8fd.png "I=\pi+[\sin x]_0^\pi")
