Welcome Back Review (Solutions)
Math 1B
1.
d ⎡ 3
2
⎤ = 3x 2 − 6x
x
−
3x
+
1
⎦
dx ⎣
2.
d
d
d
2sin x cos x ] = 2sin x ⋅ [cos x] + 2 [sin x]⋅ cos x
[
dx
dx
dx
= −2sin 2 x + 2 cos2 x
= 2(cos2 x − sin 2 x)
OR
d
d
2sin x cos x ] = [sin 2x]
[
dx
dx
= cos 2x ⋅
d
[2x]
dx
= 2 cos 2x
3.
d ⎡
ln
dx ⎣
(
)
1
d
12
sec x ⎤ =
⋅ ⎡( sec x ) ⎤
⎦
⎣
⎦
sec x dx
=
1
1
−1 2 d
⋅ ( sec x )
⋅ [sec x]
dx
sec x 2
=
1
⋅sec x ⋅ tan x
2sec x
=
tan x
2
4.
3x 3 6x 2
3
2
∫ (3x − 6x) dx = 3 − 2 + C = x − 3x + C
5.
∫ cos x e
2
sin x
= ∫ cos x ⋅ eu
dx
du
cos x
= ∫ eu du
= eu + C = esin x + C
6.
1
−1
∫ 1+ x 2 dx = tan x + C
Let u = sin x
Then
du
= cos x
dx
and dx =
du
cos x