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

13.2 Elementary integration 431
Table13.1
Theintegrals ofsome common functions.
f(x)
∫
f(x)dx
f(x)
∫
f(x)dx
k, constant kx +c
x n x n+1
+c n≠−1
n +1
x −1 = 1 x
ln|x|+c
e x e x +c
e −x −e −x +c
e ax
sinx
sinax
sin(ax +b)
cosx
cosax
e ax
a +c
−cosx+c
−cosax
+c
a
−cos(ax +b)
+c
a
sinx+c
sinax
a
+c
cos(ax +b)
tanx
tanax
tan(ax +b)
cosec(ax +b)
sec(ax +b)
sin(ax +b)
+c
a
ln|secx|+c
ln|secax|
+c
a
ln|sec(ax +b)|
+c
a
1
{ln|cosec(ax +b)
a
−cot(ax +b)|} +c
1
{ln|sec(ax +b)
a
+tan(ax +b)|} +c
cot(ax +b)
1
{ln|sin(ax +b)|} +c
a
1
√
a2 −x 2 sin −1 x a +c
1
a 2 +x 2 1
a tan−1 x a +c
Notethata,b,nandcareconstants.Whenintegratingtrigonometricfunctions,
anglesmustbeinradians.
(c) FromTable13.1,wefind ∫ −cos(ax +b)
sin(ax+b)dx =
a
andb= 2,and so
∫
−cos(3x +2)
sin(3x +2)dx = +c
3
+c.Inthiscasea = 3
(d) From Table 13.1, we find thatifkisaconstant then ∫ k dx =kx +c.Hence,
∫
5.9 dx = 5.9x +c
(e) In this example, the independent variable ist but nevertheless from Table 13.1 we
can deduce
∫
ln|sec(at +b)|
tan(at +b)dt = +c
a
Hence witha = 6 andb=−4,we obtain
∫
ln|sec(6t −4)|
tan(6t −4)dt = +c
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