Reminder on using the product rule to integrate

Reminder on using the product rule to integrate

Reduction Formulae Notation

Integration by parts formula Ex1 In = Using the By Parts Formula u = xn so = nxn–1 = e–x so integrating v = –e–x

This formula can now be used to evaluate I1, I2, I3….. i.e find I3 Find Ex2 In = In = –e–1 + nIn–1 from formula previously derived I3 = –e–1 + 3I2 I2 = –e–1 + 2I1 I1 = –e–1 + 1I0

I1 = –e–1 + 1I0 I1 = –e–1 + 1(1–e–1) = 1 – 2e–1 I2 = –e–1 + 2I1 I2 = –e–1 + 2(1 – 2e–1) = 2 – 5e–1 I3 = –e–1 + 3I2 I3 = –e–1 + 3(2 – 5e–1) = 6 – 16e–1 • 6 – 16e–1

Ex3 In = Hence find Show that In = Now use the by parts rule Integration by parts formula u = sinn–1x = (n – 1)sin n–2x cosx Using the By Parts Formula = sinx so integrating v = –cosx

Integration by parts formula u = sinn–1x = (n – 1)sin n–2x cosx = sinx so integrating v = –cosx

Multiply out and rearrange nIn = (n – 1)In–2 We need So

Reduction Formula Tricks Now integrate the sec2x tann–2x using inspection Same idea for Now integrate the 1 and differentiate the ln expression

Now integrate the sinx and differentiate the sinn–1x. Remember sin2x + cos2x = 1 Same idea for Look at the reduction formula answer for a clue as to which term has been integrated and which has been differentiated.

Reminder on using the product rule to integrate