Mathematics Learning Centre Integration: Using the chain rule in reverse Mary Barnes c 999 University of Syney
Mathematics Learning Centre, University of Syney Using the Chain Rule in Reverse Recall that the Chain Rule is use to ifferentiate composite functions such as cos( 3 +),e 2 2, (2 2 +3), ln(3+). (The Chain Rule is sometimes calle the Composite Functions Rule or Function of a Function Rule.) If we observe carefully the answers we obtain when we use the chain rule, we can learn to recognise when a function has this form, an so iscover how to integrate such functions. Remember that, if y = f(u) an u = g() so that y = f(g()), (a composite function) y then = y u u. Using function notation, this can be written as y = f (g()) g (). In this epression, f (g()) is another way of writing y u an g () is another way of writing u where u = g(). This last form is the one you shoul learn to recognise. where y = f(u) an u = g() Eamples By ifferentiating the following functions, write own the corresponing statement for integration. i. sin 3 ii. (2 +) 7 iii. e 2 Solution i sin 3 = cos 3 3, so cos 3 3 = sin 3 + c. ii iii (2 +)7 = 7(2 +) 6 2, so ( e 2) = e 2 2, so 7(2 +)6 2 = (2 +) 7 + c. e 2 2 = e 2 + c.
Mathematics Learning Centre, University of Syney 2 Eercises. Differentiate each of the following functions, an then rewrite each result in the form of a statement about integration. i (2 4) 3 ii sin π iii e 3 5 iv ln(2 ) v 5 3 vi tan 5 vii ( 5 ) 4 viii sin( 3 ) i e cos 5 i tan ( 2 +) ii ln(sin ) The net step is to learn to recognise when a function has the forms f (g()) g (), that is, when it is the erivative of a composite function. Look back at each of the integration statements above. In every case, the function being integrate is the prouct of two functions: one is a composite function, an the other is the erivative of the inner function in the composite. You can think of it as the erivative of what s insie the brackets. Note that in some cases, this erivative is a constant. For eample, consier e 3 3. We can write e 3 as a composite function. 3 is the erivative of 3 i.e. the erivative of what s insie the brackets in e (3). This is in the form f (g()) g () with u = g() =3, an f (u) =e u. Using the chain rule in reverse, since (f(g())) = f (g()) g () wehave f (g()) g () = f(g()) + c. In this case e 3 3 = e 3 + c. If you have any oubts about this, it is easy to check if you are right: ifferentiate your answer! Now let s try another: cos( 2 +5) 2. cos( 2 + 5) is a composite function. 2 is the erivative of 2 + 5, i.e. the erivative of what s insie the brackets.
Mathematics Learning Centre, University of Syney 3 So this is in the form f (g()) g () with u = g() = 2 + 5 an f (u) = cos u. Recall that if f (u) = cos u, f(u) = sin u. So, cos( 2 +5) 2 = sin( 2 +5)+c. Again, check that this is correct, by ifferentiating. People sometimes ask Where i the 2 go?. The answer is, Back where it came from. If we ifferentiate sin( 2 + 5) we get cos( 2 +5) 2. So when we integrate cos( 2 +5) 2 we get sin( 2 + 5). Eamples Each of the following functions is in the form f (g()) g (). Ientify f (u) an u = g() an hence fin an inefinite integral of the function. i. (3 2 ) 4 6 ii. sin( ) 2 Solutions i. (3 2 ) 4 6 is a prouct of (3 2 ) 4 an 6. Clearly (3 2 ) 4 is the composite function f (g()). So g() shoul be 3 2. 6 is the other part. This shoul be the erivative of what s insie the brackets i.e. 3 2, an clearly, this is the case: (32 ) = 6. So, u = g() =3 2 an f (u) =u 4 giving f (g()) g () =(3 2 ) 4 6. If f (u) =u 4, f(u) = 5 u5. So, using the rule f (g()) g () = f(g()) + c we conclue (3 3 ) 4 6 = 5 (32 ) 5 + c. You shoul ifferentiate this answer immeiately an check that you get back the function you began with.
Mathematics Learning Centre, University of Syney 4 ii. sin( ) 2 This is a prouct of sin( ) an 2. Clearly sin( ) is a composite function. The part insie the brackets is, so we woul like this to be g(). The other factor 2 ought to be g (). Let s check if this is the case: g() = = 2,sog () = 2 2 = = 2 2 2. So we re right! Thus u = g() = an f (u) = sin u giving f (g()) g () = sin( ) 2. Now, if f (u) = sin u, f(u) = cos u. So using the rule f (g()) g () = f(g()) + c we conclue sin( ) 2 = cos( )+c. Again, check immeiately by ifferentiating the answer. Note: The eplanations given here are fairly lengthy, to help you to unerstan what we re oing. Once you have graspe the iea, you will be able to o these very quickly, without neeing to write own any eplanation. Eample Integrate sin 3 cos. Solution sin 3 cos = (sin ) 3 cos. So u = g() = sin with g () = cos. An f (u) =u 3 giving f(u) = 4 u4. Hence sin 3 cos = 4 (sin )4 + c = 4 sin4 + c.
Mathematics Learning Centre, University of Syney 5 Eercises.2 Each of the following functions is in the form f (g()) g (). Ientify f (u) an u = g() an hence fin an inefinite integral of the function. i 3 3 ii 2 + 2 iii (ln )2 iv e 2+4 2 v sin( 3 ) 3 2 vi cos ( ) π π 2 2 vii (7 8) 2 7 viii sin(ln ) i ( ) cos sin e tan sec 2 i e 3 3 2 ii sec 2 (5 3) 5 iii (2 ) 3 2 iv sin cos The final step in learning to use this process is to be able to recognise when a function is not quite in the correct form but can be put into the correct form by minor changes. For eample, we try to calculate 3 4 +. We notice that 4 + is a composite function, so we woul like to have u = g() = 4 +. But this woul mean g () =4 3, an the integran (i.e. the function we are trying to integrate) only has 3. However, we can easily make it 4 3, as follows: 4 + 4 3. 3 4 + = 4 Note: The 4 an the 4 cancel with each other, so the epression is not change. So u = g() = 4 +, g () =4 3 An f (u) =u 2 f(u) = 2u 3 2 3 So, 3 4 + = 4 + 4 2 = 4 4 2 ( 4 + ) 3 2 + c. 3 Note: We may only insert constants in this way, not variables. We cannot for eample evaluate e 2 by writing e 2 2, because 2 the in front of the integral sign oes not cancel with the which has been inserte in the integran. This integral cannot, in fact, be evaluate in terms of elementary functions.
Mathematics Learning Centre, University of Syney 6 The eample above illustrates one of the ifficulties with integration: many seemingly simple functions cannot be integrate without inventing new functions to epress the integrals. There is no set of rules which we can apply which will tell us how to integrate any function. All we can o is give some techniques which will work for some functions. Eercises.3 Write the following functions in the form f (g()) g () an hence integrate them: i cos 7 ii e 2 iii 2 2 iv 2 (4 3 +3) 9 v sin( + 3) vi sin vii ( 2 ) viii e 3 i tan 6 Hint: Write tan 6 in terms of sin 6 an cos 6.
Mathematics Learning Centre, University of Syney 7 2 Solutions to eercises Eercises. i ii iii iv v vi vii viii i i ii (2 4)3 = 3 (2 4) 2 2, so (sin π) = cos π π, so (e3 5 ) = e 3 5 3, so (ln(2 )) = 2 2, so ( 5 3 ) = (5 3) 5, so 2 (tan 5) = sec2 5 5, so ((5 ) 4 ) = 4( 5 ) 3 5 4, so (sin 3 ) = cos( 3 ) 3 2, so (e ) = e 2 2, so (cos5 ) = 5 cos 4 ( sin ), so (tan(2 + )) = sec 2 ( 2 +) 2, so (ln(sin )) = cos, sin so Eercises.2 3(2 4) 2 2 = (2 4) 3 + c. cos π π = sin π + c. e 3 5 3 = e 3 5 + c. 2 = ln(2 ) + c. 2 (5 3) 5 = 2 5 3 + c. sec 2 5 5 = tan 5 + c. 4( 5 ) 3 5 4 = ( 5 ) 4 + c. cos( 3 ) 3 2 = sin( 3 )+c. e 2 2 = e + c. 5 cos 4 ( sin ) = cos 5 + c. sec 2 ( 2 +) 2 = tan( 2 +)+c. cos = ln(sin )+c. sin (Before you rea these solutions, check your work by ifferentiating your answer.) i. 3 = ln(3 ) + c. 3 u = g() = 3 so g () = 3 f (u) so f(u) = lnu = u ii. 2 + 2 = 2 3 (2 +)3 2 + c.
Mathematics Learning Centre, University of Syney 8 u = g() = 2 + f (u) = u so g () = 2 so f(u) = 2 3 u 3 2 iii. (ln ) 2 = 3 (ln )3 + c. u = g() = ln so g () = f (u) = u 2 so f(u) = 3 u3 iv. v. vi. vii. e 2+4 2 = e 2+4 + c. u = g() = 2 +4 so g () = 2 f (u) = e u so f(u) = e u sin( 3 ) 3 2 = cos( 3 )+c. u = g() = 3 f (u) cos( π 2 ) π 2 u = g() f (u) = sin u = sin(π 2 )+c. = π 2 = cos u so g () = 3 2 so f(u) so g () = π 2 so f(u) (7 8) 2 7 = 3 (7 8)3 + c. = cos u = sin u u = g() = 7 8 so g () = 7 f (u) = u 2 so f(u) = 3 u3 viii. sin(ln ) = cos(ln )+c. u = g() = ln so g () = f (u) = sin u so f(u) = cos u i. cos = ln(sin )+c. sin u = g() = sin so g () = cos f (u) = u so f(u) = lnu. e tan sec 2 = e tan + c. u = g() = tan so g () = sec2 f (u) = e u so f(u) = e u
Mathematics Learning Centre, University of Syney 9 so g () = 32 i. e 3 3 2 = e 3 + c. u = g() = 3 f (u) = e u so f(u) = e u ii. sec 2 (5 3) 5 = tan(5 3) + c. u = g() = 5 3 so g () = 5 f (u) = sec 2 u so f(u) = tan u iii. (2 ) 3 3 2 = 4 (2 ) 4 3 + c. u = g() = 2 so g () = 2 f (u) = u 3 so f(u) = 3u 4 3 4 2 iv. sin cos = 3 (sin ) 3 2 + c. u = g() = sin so g () = cos f (u) = u so f(u) = 2u 3 2 3 Eercises.3 (Before reaing the solutions, check all your answers by ifferentiating!) i. cos 7 = cos 7 7 = sin 7 + c. 7 7 u = g() =7, g () =7 f (u) = cos u so f(u) = sin u ii. e 2 = e 2 2 = + c. 2 2 e2 u = g() = 2, g () =2 f (u) = e u so f(u) =e u iii. iv. 2 = 2 4 2 2 ( 4) = 4 ln( 22 )+c. u = g() = 2 2, g () = 4 f (u) = so f(u) =lnu u 2 (4 3 +3) 9 = (4 3 +3) 9 2 2 = 2 2 0 (43 +3) 0 + c = 20 (43 +3) 0 + c. u = g() =4 3 +3, g () =2 2 f (u) = u 9 so f(u) = 0 u0
Mathematics Learning Centre, University of Syney 0 v. sin( + 3) = 3 u = g() =+3, g () =3 sin( + 3) 3 = cos( + 3)+c. 3 f (u) = sin u so f(u) = cos u vi. sin =2 sin u = g() =, g () = 2 f (u) = sin u so f(u) = cos u 2 = 2 cos + c. vii. = 2 2 2 ( 2) = 2 2( 2 ) 2 + c = ( 2 ) 2 + c. u = g() = 2, g () = 2 f (u) = u so f(u) =2u 2 viii. e 3 = e 3 3 = 3 3 e3 + c. u = g() =3, g () =3 f (u) = e u so f(u) =e u i. tan 6 = sin 6 cos 6 = 6 u = g() = cos 6, g () = 6 sin 6 cos 6 6sin 6 = ln(cos 6)+c. 6 f (u) = so f(u) =lnu u