Integration by substitution Try integrating: x 5 dx 5 Answer: Hint: Try a guess and differentiate it. x 5 6 6 c That was easy to guess, but lets try it with substitution: x 5 dx 5 Let u (x 5) u dx 5 (the dx is a problem) du u x 5, 1,dx du dx So, So we have: u6 (x 5)6 u du 6 c 6 c 5 That’s a simple one …. now let’s try a harder one …….. Try this integral: 2 4 x(x 3) dx 2 u x 3 so, du 2x dx Re-write your integral: and du dx 2x du xu 2x 1 u4 du 2 4 5 Integrate and then substitute: u c 10 x2 3 5 10 c Now try some of your own (about 10 minutes). 7 6 x 4 1. x 4 dx 2. 3. 4. 5. 6x3x x 2 2 3 x 5 dx 2 xcos(x 3)dx x x e 2 3 3x 8 dx 3 4 dx 7 2 c 8 4 c 4 3 2 2x 3 5 9 sin x 2 3 2 c c x2 4 e c 3 What about definite integration? x 3 dx 1 5 0 u x 3 Take the upper and lower values of x from the integral and find the corresponding values of u. x 1,u 2 x 0,u 3 Now re-write the integral in terms of u, as you did before. u6 2 3 u du 6 3 2 5 64 729 665 6 6 6 Confirm this result by using your Graphics Calculator. Now some of your own (about 10 minutes). 2 x2 1. xe dx 1 25.9 x cosx 2 1dx 0.0787 2. 3. 4 0 3 1 x dx 2 x 1 0.805 One step beyond…. Try this one:- x dx 2 x 3 du (too easy!) u x 3, 1,dx du dx u x3 x 2 du Something wrong? - That x, but ……. x u 3 u u 3 du 2 u 1 3 u u2 du 3 ln u c u 3 ln x 3 c x 3