Quiz 7. Time: 15 minutes
KEY
Name:
1
Math 142, Calculus II, Fall 2016
Section:
ID#:
(i) (1pt) Which of the following is a correct identity.
A. sec2 x = 1 − tan2 x. B. tan2 x = sec2 x − 1.
C. cot2 x = csc2 x + 1.
D. sec2 x − tan2 x = −1.
(ii) (1pt) Which of the following is a correct identity.
A. cos2 x = 1 − 2 sin(2x). B. sin2 x = 12 (1 − cos(2x)).
C. cos2 x = 21 (1 − cos(2x)).
D. B & C.
(iii) (1pt) Which of the following is a correct identity.
A. cos2 (2x) =
2
D. sin (2x) =
1
2 (1
B. cos2 (2x) =
+ cos(4x)).
1
4 (1
+ cos(4x)).
C. sin2 (2x) =
1
2 (1
+ cos(4x)).
1
4 (1
+ cos(4x)).
Z
(iv) (1pt) Consider the integral
tan3 x sec x dx. To calculate this integral, we should use the substitution
A. u = sec x.
D. u = tan2 x.
C. u = sec2 x.
B. u = tan x.
Solution:
We split tan3 x into tan2 x tan x and then use the identity in part (i)
Z
Z
Z
3
2
tan x sec x dx = tan x sec x tan x dx = (sec2 x − 1) sec x tan x dx
Let u = sec x, then du = sec x tan x dx and so the integral becomes
Z
sec3 x
u3
−u+C =
− sec x + C.
(u2 − 1) du =
3
3
J
Z
(v) (1pt) Consider the integral
A. u = sec x. B. u = tan x.
tan5 x sec4 x dx. To calculate this integral, we should use the substitution
C. u = sec2 x.
D. u = tan2 x.
Solution:
We split sec4 x into sec2 x sec2 x and then use the identity in part (i)
Z
Z
Z
Z
tan5 x sec4 x dx = tan5 x sec2 x sec2 x dx = tan5 x(1 + tan2 x) sec2 x dx = (tan5 x + tan7 x) sec2 x dx
Let u = tan x, then du = sec2 x dx and so the integral becomes
Z
u6
u8
tan6 x tan8 x
(u5 + u7 ) du =
+
+C =
+
+ C.
6
8
6
8
J
Z
(vi) Consider the integral
dx
.
(x2 + 4)2
(vi1) (1pt) To calculate this integral, we should use the substitution
A. x = 2 sec θ. B. x = 2 tan θ.
C. x = 4 sin θ.
D. x = 2 sin θ.
(vi2) (1pt) With the correct choice of substitution, the integral can be written as
Z
Z
Z
Z
2
2
2
1
1
1
1
A. 8 sin θ dθ. B. 8 cos θ dθ.
C. 4 sin θ dθ.
D. 4 cos2 θ dθ.
Quiz 7. Time: 15 minutes
Math 142, Calculus II, Fall 2016
2
Solution:
dx = 2 sec2 θ dθ and
(x2 + 4)2 = ([2 tan θ]2 + 4)2 = (4 tan2 θ + 4)2 = (4[tan2 θ + 1])2 = (4 sec2 θ)2 = 16 sec4 θ. Thus,
Z
dx
=
(x2 + 4)2
Z
2 sec2 θ dθ
1
=
16 sec4 θ
8
Z
dθ
1
=
sec2 θ
8
Z
cos2 θ dθ
J
2
Z
x
√
dx is equivalent to
(vii) (1pt) The integral
1 − x2
Z
Z
Z
sin2 θ dθ.
C.
sin θ cos θ dθ.
A.
sin θ dθ. B.
D. None of these choices.
Solution:
p
√
√
Let x = sin θ, then dx = cos θ dθ and 1 − x2 = 1 − sin2 θ = cos2 θ = cos θ. Plugging all these in the
integral we have
Z
Z
Z
x2
sin2 θ
√
dx =
(cos θ dθ) = sin2 θ dθ.
cos θ
1 − x2
To calculate this integral we should use the identity in part (ii) sin2 θ =
Z
1 cos(2θ)
−
to have:
2
2
Z
θ sin(2θ)
1 cos(2θ)
−
dθ = −
+C
2
2
2
4
θ 2 sin θ cos θ
= −
+C
2
4 √
sin−1 (x) x 1 − x2
=
−
+C
2
2
√
where in the last line we used the fact that x = sin θ and cos θ = 1 − x2 .
sin2 θ dθ =
J
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