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

464 Chapter 14 Techniques of integration
Example14.8 Evaluate
∫ 2
1
sint cos 2 tdt.
Solution Putz = cost so that dz = −sint, that is sintdt = −dz. Whent = 1,z = cos1; when
dt
t = 2,z = cos2.Hence
∫ 2
∫ cos2
[ ] z
sint cos 2 tdt =− z 2 3 cos2
dz=−
3
∫
Example14.9 Find
1
e
tanx
cos 2 x dx.
cos1
cos1
= cos3 1 −cos 3 2
3
Solution Putz = tanx. Then dz
dx = sec2 x,dz =
dx
cos 2 x . Hence,
∫ e
tanx
∫
cos 2 x dx = e z dz=e z +c=e tanx +c
∫
Example14.10 Find
Integration by substitution allows functions ofthe form df/dx
f
3x 2 +1
x 3 +x+2 dx.
= 0.0766
tobe integrated.
Solution Putz =x 3 +x+2,then dz
dx = 3x2 +1,thatisdz = (3x 2 +1)dx. Hence,
∫
3x 2 ∫
+1 dz
x 3 +x+2 dx = z =ln|z|+c=ln|x3 +x+2|+c
∫ df/dx
Example14.11 Find dx.
f
Solution Putz = f.Then dz
dx = df
dx , thatis
dz = df
dx dx.
Hence,
∫ df/dx
f
dx =
∫ dz
z =ln|z|+c=ln|f|+c
and so
∫
df/dx
f
dx=ln|f|+c