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

Therefore,
∫
∫ 1
sinmtsinnt dt = {cos(m −n)t −cos(m +n)t}dt
2
= 1 { }
sin(m −n)t sin(m +n)t
− +c
2 m −n m +n
13.2 Elementary integration 439
EXERCISES13.2
1 Integratethe following expressions usingTable 13.1:
1 1
(e) √ (f)
(a) x 10 (b) 9 (c) x 1.5
0.25 −x 2 0.01 + v 2
(d) √ 1 1
1
(g)
t (e) (f) −3.2
10 6 +r 2 (h)
10+t 2
z
1 1
1 1
(g)
x 3 (h) √ (i) (x 2 ) 3
(i) √ (j)
t 2−x 2 1
(j) x −2 (k) t 1/3
9 +x2
5 Integratethe followingexpressions:
2 Integratethe following expressions usingTable 13.1:
(a) e 5x (b) e 6x (c) e −3t (a)3+x+ 1 x
1
(d)
e x (e) e 0.5t (f) e z/3 (b)e 2x −e −2x
1
(g)
e 4t (h) e −2.5x
(c)2sin3x +cos3x
( )
t
(d)sec(2t + π) +cot
3 Integratethe following expressions usingTable 13.1:
2 − π
( )
(a) sin4x (b) sin9t
( )
( )
t
(e)tan +cosec(3t − π)
x 2t
2
(c) sin (d) sin
2 5
(e) cos7x (f) cos(−3x)
(f) sinx + x 3 + 1 e
( )
x
5t
1
(g) cos (h) tan9x
(g)
3
cos(3x)
( ) 2
(i) cosec2x (j) sec5t
(h) t + 1 (k) cot8y (l) cos(5t +1)
t
(m) tan(3x +4) (n) sin(3t − π)
1
( ) (i)
x
3e 2x
(o) cosec(5z +2) (p) sec
2 +1 (j) tan(4t −3) +2sin(−t −1)
( )
2t
(k)1+2cot3x
(q) sin
3 −1 (r) cot(5 −x)
( ) ( )
t t
(l) sin −3cos
(s) cosec(π −2x)
2 2
4 UseTable 13.1 to integrate the following expressions: (m)(t −2) 2
1 1
(n)3e −t −e −t/2
(a)
1+x 2 (b) √
1−x 2
(o)7 −7x 6 +e −x
1 1
(c) √ (d)
(p) (k +t) 2 k constant
4−x 2 9+z 2 (q)ksint −coskt k constant