First year: (I) Integration by parts Formula and example It is relatively easy to derive the formula: u( x)v( x)' u' ( x)v( x) u( x)v' ( x) u' ( x)v( x) u( x)v( x)'u( x)v' ( x). (product rule) Now, integrate on both sides: u ' ( x)v( x)dx u ( x)v( x)' dx u ( x)v' ( x)dx and finally u ' ( x)v( x)dx u ( x)v( x) u ( x)v' ( x)dx. The formula above is the formula for integration by parts. With the alternative notation for the derivative, the formula can be written as follows: du ( x) dv( x) v( x)dx u ( x)v( x) u ( x)dx. dx dx Example: Calculate x sin( x)dx . The trick consists in choosing wisely what your functions u and v will be. Here we have to choose u ' ( x) sin( x) and v( x) x . Those choices lead to u ( x) cos( x) and v' ( x) 1, so that, applying the formula, we end up with x sin( x)dx x cos( x) (1 cos( x))dx x cos( x) sin( x). Had we chosen differently, we would have made the problem worse: take u ' ( x) x and v( x) sin( x) this time. This choice leads to u ( x) x2 , v' ( x) cos( x) 2 x sin( x)dx x2 1 sin( x) x 2 cos( x)dx. 2 2 This is mathematically correct, but the integral on the right-hand side is even more complicated than the one we began with. We have only managed to make the problem worse. (II) Worked examples and exercises (i) Calculate ln( x)dx. Here, it is not obvious that integration by parts is needed (until the other possibilities have been exhausted, that is). We have ln( x)dx 1 ln( x)dx (it sometimes pays to state the obvious). Let’s set u ' ( x) 1 and v( x) ln( x). We then have u ( x) x and v ' ( x ) 1 , so that x 1 ln( x)dx x ln( x) x x dx x ln( x) x. (ii) Calculate x 5 ln( x) dx . Similar to the above. Give it a go… (iii) Calculate x 2 sin( x)dx. Here, set u ' ( x) sin( x) and v( x) x 2 . We have u ( x) cos( x) and v' ( x) 2 x so that x 2 sin( x)dx x 2 cos( x) 2 x cos( x)dx. It is not a dead end, you just have to use integration by parts again on the integral on the right-hand side. Finish it… (iv) Calculate cos( x)e x dx. This example is a classic, but could be quite tricky the first time you do it. Set u ' ( x) cos( x) and v( x) e x . Then u ( x) sin( x) and v' ( x) e x , so that cos( x)e dx sin( x)e sin( x)e dx. x x x It does not look like much of an improvement, but this time we persevere and integrate the right-hand side by parts again: u ' ( x) sin( x) u ( x) cos( x) , so that v( x) e x v' ( x) e x sin( x)e dx cos( x)e cos( x)e dx. x x x Hence cos( x)e dx sin( x)e x x cos( x)e x cos( x)e x dx so that cos( x)e dx 2 sin( x)e x 1 (v) Calculate sin( x)e x dx. (vi) Calculate 2 x cos( x / 2)dx. x cos( x)e x .