M408D SPRING 2015 HOMEWORK #2 (1) - cheatham (sc36975) HW02...

cheatham (sc36975) – HW02 – um – (53890) 1
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001 10.0 points
Evaluate the integral
keywords: Stewart5e, indefinite integral,
powers of sin, powers of cos, trig substitu-
tion,
002 10.0 points

cheatham (sc36975) – HW02 – um – (53890) 2
which as the graph shows can in turn be writ-
ten as
I =
π
0
2 sin
3
( x ) dx −
3 π / 2
π
2 sin
3
( x ) dx .
Since
sin
2
( x ) = 1 − cos
2
( x ) ,
we thus see that
I =
π
0
−
3 π / 2
π
2 sin( x )(1 − cos
2
( x )) dx .
To evaluate these integrals, set u = cos( x ).
For then
du = − sin( x ) dx ,
in which case
π
0
2 sin( x )(1 − cos
2
( x )) dx
= − 2
− 1
1
(1 − u
2
) du = 2
1
− 1
(1 − u
2
) du
= 2 u −
1
3
u
3
1
− 1
=
8
3
,
while
3 π / 2
π
2 sin( x )(1 − cos
2
( x )) dx
= − 2
0
− 1
(1 − u
2
) du
= − 2 u −
1
3
u
3
0
− 1
= −
4
3
.
Consequently, the shaded region has
area = 4 .
keywords:
003 10.0 points
Evaluate the definite integral
I =
π / 4
0
4 cos x − 2 sin x
cos
3
x
dx .
1. I =
5
2
2. I = 2
3. I =
7
2
4. I = 4
5. I = 3 correct
Explanation:
After division
4 cos x − 2 sin x
cos 3 x
= 4 sec
2
x − 2 tan x sec
2
x
= (4 − 2 tan x ) sec
2
x .
Thus
I =
π / 4
0
(4 − 2 tan x ) sec
2
x dx .
Let u = tan x ; then
du = sec
2
x dx
so
I =
1
0
(4 − 2 u ) du = 4 u − u
2
1
0
.
Consequently,
I = 3 .
004 10.0 points
Evaluate the integral
I = 2
2 π / 3
0
1 + cos( θ ) d θ .
Hint: use a double angle formula to express
1 + cos( θ ) in terms of cos
2
( θ / 2).
1. I = 2
√
2

cheatham (sc36975) – HW02 – um – (53890) 3
2. I = 4
√
6
3. I = 4
√
2
4. I = 4
5. I = 2
6. I = 2
√
6 correct
Explanation:
By a double angle formula for cos(2 x ),
cos(2 x ) = 2 cos
2
( x ) − 1 ,
so, taking x = θ / 2, we see that
1 + cos( θ ) = 2 cos
2
θ
2
=
√
2 cos
θ
2
when cos( θ / 2) ≥ 0, hence for 0 ≤ θ ≤ π .
Thus
I = 2
√
2
2 π / 3
0
cos
θ
2
d θ = 4
√
2 sin
θ
2
2 π / 3
0
.