How to integrate some special products of two functions Integration by parts How to integrate some special products of two functions April 2011 纪光 - 北京 景山学校

Problem Not all functions can be integrated by using simple derivative formulas backwards … Some functions look like simple products but cannot be integrated directly. Ex. f(x) = x .sin x but … u’.v’≠ (u.v)’ So what ?!? April 2011 纪光 - 北京 景山学校

Reminder 1 Derivative of composite functions if g(x) = f[u(x)] and u and f have derivatives, then g’(x) = f’[u(x)] . u’(x) hence April 2011 纪光 - 北京 景山学校

Examples of integrals products of composite functions April 2011 纪光 - 北京 景山学校

Formulas of integrals of products made of composite functions April 2011 纪光 - 北京 景山学校

Reminder 2 Derivative of the Product of 2 functions If u and v have derivatives u’ and v’, then (u.v)’ = u’.v + u.v’ u.v’ = (u.v)’ - u’.v hence by integration of both sides April 2011 纪光 - 北京 景山学校

Example 1 Integration by parts u’ =(x)’ = 1 u = x sin x = (- cos x)’ u = 1st part (to derive) v’ = 2nd part (to integrate) u’ =(x)’ = 1 u = x sin x = (- cos x)’ v’ = sin x u’v = 1.(- cos x) v = - cos x April 2011 纪光 - 北京 景山学校

Example 1 Integration by parts u = x  u’ = 1 v’ = sin x  v = - cos x April 2011 纪光 - 北京 景山学校

Example 2 Integration by parts u’ =(x2)’ = 2x u = x2 cos x = (sin x)’ 1st part (to derive) 2nd part (to integrate) u’ =(x2)’ = 2x u = x2 cos x = (sin x)’ v’ = cos x u’v = 2x.sin x v = sin x April 2011 纪光 - 北京 景山学校

Example 2 Integration by parts u = x2  u’ = 2x v’ = cos x  v = sin x April 2011 纪光 - 北京 景山学校

Example 3 Integration by parts u’ =(ln x)’ = u = ln x x = =>v = . 1st part (to derive) 2nd part (to integrate) u’ =(ln x)’ = u = ln x x = =>v = . v’ = x u’v = April 2011 纪光 - 北京 景山学校

Example 3 Integration by parts u = ln x  u’ = v’ = x  v = April 2011 纪光 - 北京 景山学校

Example 4 Integration by parts u’ =(ln x)’ = u = ln x 1st part (to derive) 2nd part (to integrate) u’ =(ln x)’ = u = ln x 1 = (x)’=> v = v’ = 1 u’v = 1 April 2011 纪光 - 北京 景山学校

Example 4 Integration by parts u = ln x  u’ = v’ = 1  v = x April 2011 纪光 - 北京 景山学校

Example 5 Integration by parts u’ =(ln x)’ = u = ln x 1st part (to derive) 2nd part (to integrate) u’ =(ln x)’ = u = ln x ln x = (x ln x – x)’ v’ = ln x u’v = ln x - 1 v = x ln x – x April 2011 纪光 - 北京 景山学校

Example 5 Integration by parts u = ln x  u’ = v’ = ln x  v = x.ln x - x April 2011 纪光 - 北京 景山学校

Problem I Integration by parts Use the IBP formula to calculate : exp(exp(1) + 1) - exp(exp(1)) April 2011 纪光 - 北京 景山学校

Problem II Integration by parts Use the IBP formula to calculate : April 2011 纪光 - 北京 景山学校

Problem III Integration by parts Let , for any Integer n ≥ 0 : Use the IBP formula to prove that : Then find I0, I1, I2 April 2011 纪光 - 北京 景山学校

Problem IV Integration by parts Use the IBP formula to prove that : Let , for any Integer n ≥ 0 : [Wallis Integral] Problem IV Use the IBP formula to prove that : Calculate I0 and I1 Find a short formula for I2n Find a short formula for I2n+1 April 2011 纪光 - 北京 景山学校

Problem V Integration by parts Use twice the IBP formula to calculate April 2011 纪光 - 北京 景山学校

Problem VI Integration by parts Use twice the IBP formula to calculate Anwser : I = April 2011 纪光 - 北京 景山学校

Problem VII Integration by parts Calculate I + J Use IBP to calculate I – J Find I and J Anwser : I = April 2011 纪光 - 北京 景山学校

Problem VIII Integration by parts Calculate I1 Find a relationship between In and In-1 (n ≥ 1) Show that Anwser : I = April 2011 纪光 - 北京 景山学校