Answer Key for Quiz 3 (section A)
1. If we substitute u=√x=x1
2, then du =1
2x−1
2dx =dx
2√x, so that 2 du =dx
√x. If x= 0 then u=√0 = 0,
and if x=π2
4then u=rπ2
4=π
2. Therefore
Zπ2
4
0
cos √x
√xdx =Zπ
2
0
(cos u)(2 du) = 2 sin u|
π
2
0= 2 ³sin π
2−sin 0´= 2 −0 = 2.
2. Formula 8 in the table applies to this integral with a=−2 and b= 3, so we have
Z∞
0
e−2xsin 3x dx =1
(−2)2+ 32e−2x[−2 sin 3x−3 cos 3x]¯
¯
∞
0
=−1
13 e−2x[2 sin 3x+ 3 cos 3x]¯
¯
∞
0.(i)
Although cos ∞and sin ∞are not well-defined, since cos xand sin xare never large and e−2xgoes to zero
very fast as x→ ∞ we have
(ii) lim
x→∞
e−2xcos 3x= 0 = lim
x→∞
e−2xsin 3x.
We also have
(iii) e−2·0cos (3 ·0) = 1 ·1 = 1 and e−2·0sin (3 ·0) = 1 ·0 = 0.
Using (ii) and (iii) in (i) we get
Z∞
0
e−2xsin 3x dx =−1
13 {[0 + 0] −[0 + 3]}=3
13.
Alternatively we could integrate by parts twice. If we take u=e−2xand dv = sin 3x dx then du =−2e−2xdx
and v=−1
3cos 3xand we have
(iv) Z∞
0
e−2xsin 3x dx =−1
3e−2xcos 3x¯
¯
∞
0−2
3Z∞
0
e−2xcos 3x dx.
Or we could take u= sin 3xand dv =e−2xdx, in which case du = 3 cos 3x dx and v=−1
2e−2xand we have
(v) Z∞
0
e−2xsin 3x dx =−1
2e−2xsin 3x¯
¯
∞
0+3
2Z∞
0
e−2xcos 3x dx.
Using (ii) and (iii), (iv) becomes
(vi) Z∞
0
e−2xsin 3x dx =−1
3(0 −1) −2
3Z∞
0
e−2xcos 3x dx =1
3−2
3Z∞
0
e−2xcos 3x dx
and (v) becomes
(vii) Z∞
0
e−2xsin 3x dx =−1
2(0 −0) + 3
2Z∞
0
e−2xcos 3x dx =3
2Z∞
0
e−2xcos 3x dx.
