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**Unformatted text preview: **Math 16b
Thomas Scanlon Autumn 2010 Thomas Scanlon Math 16b Chain Rule Recalled or df dg d (f (g (x )) = | dx du u=g (x ) dx
(f ◦ g ) (x ) = f (g (x ))g (x ) Thomas Scanlon Math 16b Inverting the Chain Rule: Integration by Substitution f (g (x ))g (x )dx = f (g (x )) + C Thomas Scanlon Math 16b Formalism of Integration by Substitution
Often, one writes the substitution as u = g (x )
and du = g (x )dx f (u )du = f (g (x ))g (x )dx and attempts to write the integrand as One is then charged with integrating f (u )du = F (u ) + C
We conclude that f (g (x ))g (x )dx =
Thomas Scanlon Math 16b f (u )du = F (u ) + C = F (g (x )) + C An integral revisited
Making the substitution u = x 2 + 1, so that du = 2xdx we have 1 2x x 2 + 1dx 2
√ x x 2 + 1dx =
= 1 2 udu Writing √ u = u 2 , we nd
√ 1 udu = u 2 + C
3 12 (x + 1) 2 + C 3 2 3 3 Thus, x x 2 + 1dx = Thomas Scanlon Math 16b Example
Integrate: xe x dx
Set u = x 2 . Then du = 2xdx . 2 xe x dx 2 = 1 2 e x 2xdx 2 1 = e u du 2 1u e +C = 2 1 x2 = e +C 2 Thomas Scanlon Math 16b Another Example
Integrate sin(x ) cos(x )dx Set u = sin(x ) so that du = cos(x )dx . sin(x ) cos(x )dx =
= udu 12 u +C 2 12 = sin (x ) + C 2 Thomas Scanlon Math 16b A third example
Integrate ln(x )
1 Set u = ln(x ) so that du = x dx . x dx ln(x ) x dx = = = √ udu u 2 du 1 3 22 u +C 3 3 2 = (ln(x )) 2 + C 3
Thomas Scanlon Math 16b A nal example
Integrate sin(x ) dx cos3 (x ) Set u = tan(x ). Then du = sec2 (x )dx . sin(x ) dx = cos3 (x )
= = =
Thomas Scanlon Math 16b sin(x ) 1 2 dx cos(x ) cos(x ) tan(x )sec2 (x )dx udu
12 1 u + C = tan2 (x ) + C 2 2 An alternate solution
Set v = cos(x ) so that dv = − sin(x )dx . sin(x ) dx = cos3 (x )
=− = 1 sin(x )dx cos(x )3 v −3 dv 1 −2 v +C 2 12 sec (x ) + C = 2 Thomas Scanlon Math 16b Error? Why are these answers dierent? tan2 (x ) + 1 = sec2 (x ) 12 1 1 sec (x ) + C = tan2 (x ) + [C + ] 2 2 2 So, C = 1 +C 2 Thomas Scanlon Math 16b ...

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