082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

438 Chapter 13 Integration
Solution Powersoftrigonometricfunctions,forexamplesin 2 t,donotappearinthetableofstandardintegrals.Whatwemustattempttodoisrewritetheintegrand
toobtainastandard
form.
(a) FromTable 3.1
cos 2 t = 1+cos2t
2
(b)
and so
∫ ∫ 1+cos2t
cos 2 t dt = dt
2
∫ ∫ 1 cos2t
=
2 dt + dt
2
= t 2 + sin2t +c
4
∫ ∫
sin 2 t dt = 1 −cos 2 t dt using the trigonometric identities
∫
=
∫
1dt−
cos 2 t dt
( t
= t −
2 + sin2t )
+c
4
= t 2 − sin2t
4
+k
using linearity
using part(a)
Example13.6 Find
(a) ∫ sin2tcost dt
(b) ∫ sinmtsinnt dt, wheremandnareconstants withm ≠n
Solution (a) Usingthe identitiesinTable3.1 wefind
2sinAcosB =sin(A +B) +sin(A −B)
hence sin2tcost can bewritten 1 (sin3t +sint). Therefore,
2
∫
∫ 1
sin2tcost dt = (sin3t +sint)dt
2
= 1 ( )
−cos3t
−cost +c
2 3
= − 1 6 cos3t − 1 2 cost+c
(b) Usingthe identity2sinAsinB = cos(A −B) −cos(A +B), we find
sinmtsinnt = 1 {cos(m −n)t −cos(m +n)t}
2