A-level Mathematics: Reverse Chain Rule
                      Priority Learning
                        Revision Sheet
         Name:                      Date:
                      Aims of this revision sheet:
          • Understanding how to use the reverse chain rule.
          • Becoming familiar with common, examinable applications of the reverse chain
           rule.
          • To improve exam technique.
          • To familiarise yourself with common exam style questions.
                             The Fundamental Theorem Of Calculus
                  Let f(x) be some function of x and let f′(x) be the first derivative of f(x). Then we have,
                                             Z f′(x)dx = f(x)+c
                   What this means in words is that if we integrate the first derivative of some function, we
                                  get the original function back plus some constant.
                                        Reverse Chain Rule - Method
                     1. Make a guess at what the integral might be. Then differentiate that to see how close
                       you are.
                     2. Repeat with trial and error until the function you are differentiating, differentiates to
                       whatever the integral is.
                     3. Don’t forget your + c.
                                                  Example
                           R     4
                   Compute (x+3) dx.
                    Solution: Well we know that the power will increase by 1. So first we will try differ-
                                 5
                    entiating (x + 3) .
                                            d      5
        4
                                           dx (x+3) =5(x+3)
                    Multiplying both sides by 1 we obtain.
                                         5
                                          1 d      5
       4
                                          5dx (x+3) =(x+3)
                    Hence,
                                         d 1      5        4
                                         dx  5(x+3)   =(x+3)
                    Hence,
                                            Z (x+3)4dx= 1(x+3)5+c
                                                        5
                                                   Page 2
                                                    Example
                    Compute R(3−2x)6dx.
                     Solution: Well we know that the power will increase by 1. So first we will try differ-
                     entiating (3 − 2x)7.
                                     d (3−2x)7
 = 7(−2)(3−2x)6   by the chain rule.
                                     dx
                                                 =−14(3−2x)6
                     So we need to multiply by - 1 in order to achieve our desired result.
                                            14
                                 −1 d (3−2x)7
=(3−2x)6
                                   14dx
                                 d − 1 (3−2x)7=(3−2x)6
                                dx    14
                     Hence,
                                     Z (3−2x)6dx = − 1 (3−2x)7 +c
                                                     14
                                     Trigonometric Differentiation Reminder
                                           d    n       d      n
                                           dx (sin (x)) = dx (sin(x) )
                                                           n−1
                                                     =nsin    (x)cos(x)
                   Wemultiply by the power, then reduce the power by 1 and use the chain rule on sin(x). If
                   this is very unfamiliar please see the chain rule revision sheet.
                   This can be extended to an function of x inside the sin.
                                     d    n         d          n
                                     dx (sin (f(x))) = dx (sin(f(x)) )
                                                      ′     n−1
                                                  =nf (x)sin  (f(x))cos(f(x))
                                                    Example
                                        d   5   2 
         4  2      2
                                        dx sin (3x ) = 5(6x)sin (3x )cos(3x )
                                                           4   2      2
                                                    =30xsin (3x )cos(3x )
                                                     Page 3
                            The Reverse Chain Rule - Trigonometric Integration
                          If we have a trigonometric function involving sin and cos of certain forms it is far more
                          straightforward to use the reverse chain rule than any other method. These forms are,
                                                  Z    n
                                                    sin (x)cos(x)dx
                        Or
                                                  Z cosn(x)sin(x)dx
                        This is because (using the results from the differentiation recap),
                                                  Z    n                   1      n+1
                                                    sin (x)cos(x)dx = n+1 sin        (x)+c
                        Similarly for the second integral,
                                                  Z     n                    1      n+1
                                                    cos (x)sin(x)dx = −n+1 cos          +c
                        If you’re unconvinced, try differentiating the right hand side of both integrals.
                                                                  Example
                                   R    2
                         Compute     sin (x)cos(x)dx.
                           Solution:
                           Weincrease the power of the sin by 1 and eliminate the cos. Then differentiate to see
                           how close our guess is.
                                                           d sin3(x)
 = 3sin2(x)cos(x)
                                                           dx
                           Hence,
                                                         1 d     3   
      2
                                                         3dx sin (x) = sin (x)cos(x)
                                                        d 1     3          2
                                                       dx    3 sin (x)  =sin (x)cos(x)
                           Hence,
                                                    Z     2                1    3
                                                       sin (x)cos(x)dx = 3 sin (x)+c
                                                                    Page 4