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MAT 296-M003 Fall 2022 - Take-Home Quiz 4
9/30/22 - Due 10/7/22; 1 Week Instructor: Ng Name: Butter Toast
Instructions: This is take-home Quiz 4 on trigonometric integrals. Please work on these problems
in your Quiz 3 groups. Refer to lecture notes from Tuesday 9/28 if you get stuck. I will be coming
around to each group to check on progress. Please complete this by next Friday’s recitation 10/7.
(1) (3 points)
Z
cos2
(2x + 1)dx
Solution. Use double angle identity:
Z
cos2
(2x + 1)dx =
Z
1
2
(1 + cos(2(2x + 1)))dx
=
1
2
Z
(1 + cos(4x + 2))dx
=
1
2
x +
1
4
sin(4x + 2)
+ C,
where in the last equality we integrated term by term and used u-sub with u = 4x + 2.
(2) (3 points)
Z
sin5 x cos3 xdx
Solution. We have odd powers so we try using Pythagorean identity after factoring one of
the odd powers such that a ”squared” cosine or sine appears; a ”good” choice would be
cos3 x = cos2 x cos x and set u = sin x:
Z
sin5 x cos3 xdx =
Z
sin5 x cos2 x cos xdx
=
Z
sin5 x(1 − sin2
) cos xdx
=
Z
u
5
(1 − u
2
)du
=
Z
(u
5 − u
7
)du
=
u
6
6
−
u
8
8
+ C
=
sin6 x
6
−
sin8 x
8
+ C.
(3) (3 points)
Z
sec4 x tan6 xdx
1
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Solution. Let’s try writing ”everything” in terms of tan x; which means that we would need
to set u = tan x so that du = sec2 xdx is a term that appears when we do a substitution:
Z
sec4 x tan6 xdx =
Z
sec2 x tan6 x sec2 xdx
=
Z
(1 + tan2 x) tan6 x sec2 xdx
=
Z
(1 + u
2
)u
6
du
=
Z
(u
6 + u
8
)du
=
u
7
7
+
u
9
9
+ C
=
tan7 x
7
+
tan9
9
+ C.
(4) (3 points)
Z √
4 − x
2dx
Solution. Note the integrand is of the form √
a
2 − x
2 where a = 2; as discussed in class, we
choose the substitution x = a sin u = 2 sin u. Then dx = 2 cos udu and we have
Z √
4 − x
2dx =
Z p
4 − (2 sin u)
22 cos udu = 2 Z p
4 − 4 sin2 u cos udu
= 2 Z q
4(1 − sin2 u) cos udu
= 4 Z p
1 − sin2 u cos udu
= 4 Z √
cos2 u cos udu
= 4 Z
cos2 udu
= 4 Z
1
2
(1 + cos 2u)du
= 2 Z
(1 + cos 2u)du
= 2
u +
sin 2u
2
+ C
= 2u + sin 2u + C
= 2 arcsin x
2
+ 2 sin u cos u + C
= 2 arcsin x
2
+ 2 x
2
√
4 − x
2
2
+ C