MAC 2312
Integration Problems Mixed Together
1)
ex
∫ 1+ ex dx
ex
∫ 1+ ex dx =
1
∫ u du
(
)
= ln u + C = ln 1+ e x + C [Absolute value is not needed because 1+ e x > 0
u=1+ex →du=exdx
( )
ex
⎡Note: e 2x = e x 2 ⎤
dx
2) ∫
⎢⎣
⎥⎦
1+ e 2x
ex
ex
dx
=
∫ 1+ e2x
∫
1+ e x
( )
3)
1
∫ 1+ e
x
2
dx =
1
∫ 1+ u
2
( )
du = tan−1 u + C = tan−1 e x + C
u=ex →du=exdx
dx
⎛
1
1+ e x − e x
ex ⎞
ex
x
dx
=
⋅
dx
=
1−
dx
=
1dx
−
∫ 1+ ex
∫ 1+ ex
∫ ⎜⎝ 1+ ex ⎟⎠
∫
∫ 1+ ex dx = x − ln 1+↑ e + C =
(
)
problem one
4)
cos x
∫ 1+ sin x dx
cos x
Use direct substitution with u = 1 + sin x
∫ 1+ sin x dx = ∫
du
u
= ln u + C = ln 1+ sin x + C
u=1+sinx
du=cosxdx
5)
∫
cos x
(1+ sin x )
8
cos x
dx
Use direct substitution with u = 1 + sin x
du
∫ 1+ sin x dx = ∫ u
8
u=1+sinx
du=cosxdx
= ∫ u−8 du =
1 −7
1
1
u +C= − 7 +C= −
−7
7u
7 1+ sin x
(
)
7
+C
6)
1
∫ 1+ sin x dx
1
1
1− sin x
1− sin x
∫ 1+ sin x dx = ∫ 1+ sin x ⋅ 1− sin x dx = ∫ 1− sin2 x dx = ∫
multiply by 1−sinx
1−sinx
∫ sec
2
xdx − ∫ sec 2 x sin x dx = ∫ sec 2 x dx −
1
1− sin x
cos 2 x
∫ (1− sin x ) sec
dx =
sin x
∫ cos x ⋅ cos x dx = ∫ sec
2
2
x dx =
x dx − ∫ sec x tan x dx =
tan x − sec x + C
Alternate Solution:
1
1− sin2 x
1
dx
=
∫ 1+ sin x
∫ 1+ sin x ⋅ 1− sin2 x dx =
2
multiply by
1
∫
(1+ sin x ) (1− sin x )
1+ sin x
1−sin x
⋅
1
1− sin x
dx = ∫
dx =
2
cos x
cos2 x
1−sin2 x
sin x
1
sin x
1
dx = ∫
dx − ∫
⋅
dx = ∫ sec 2 x dx − ∫ tan x sec x dx =
2
2
cos x cos x
x
cos x
cos x
tan x − sec x + C
∫ cos
7)
2
dx − ∫
sin x
∫ 1+ sin x dx
1+ sin x − 1
1+ sin x
1
1
dx = ∫
dx − ∫
dx = ∫ 1dx − ∫
dx =
1+ sin x
1+ sin x
1+ sin x
1+ sin x
x − tan x − sec x + C = sec x − tan x + x + C
sin x
∫ 1+ sin x dx = ∫
(
8)
∫x
2
∫x
2
↑
problem 6
)
x
dx
+ 8x + 25
x
dx =
− 8x + 25
1 1
dw
2∫w
∫ (x
x
2
w=u2 +9→dw=2udu→ 1dw=udu
2
)
− 8x + 16 + 25 − 16
+ 4∫
dx =
∫
x
( x − 4)
2
+9
dx =
u+ 4
u
4
du = ∫ 2
du + ∫ 2
du =
2
+9
u +9
u +9
∫u
u=x−4→du=dx
(
du
1
u
∫ u2 +a2 = a tan−1a +C
2
1 ⎡
4
x−4
1
4
x−4
ln x − 4 + 9 ⎤ + tan−1
+ C = ln x 2 − 8x + 25 + tan−1
+C
⎢
⎥
⎦ 3
2 ⎣
3
2
3
3
(
)
)
1
1
1
u
1
4
u
du = ln w + 4 ⋅ tan−1 + C = ln u2 + 9 + tan−1 + C =
2
3
3
2
3
3
u +9
2
(
)
9)
∫x
2
∫x
2
dx
− 8x + 25
dx
=
− 8x + 25
∫ (x
dx
2
)
− 8x + 16 + 25 − 16
=
∫
(
dx
)
2
x−4 +9
∫u
=
du
+9
1
u
tan−1
3
3
=
2
du
u=x−4⇒du=dx
1
+C=
u
∫ u2 +a2 = 3 tan−1a +C
1
x−4
tan−1
+C
3
3
10)
∫
∫
(x
(x
=−
11)
∫
∫
x−4
2
− 8x + 25
x−4
2
− 8x + 25
2
dx
2
dx =
)
)
1 1
du
2 ∫ u2
(
=
)
(
)
u=x 2 −8x+25→du= 2x−8 dx→ 1du= x−4 dx
2
1 −2
1 1
u du = ⋅ u−1 + C
∫
2
2 −1
1
1
+C= −
+C
2
2u
2 x − 8x + 25
(
1
(x
2
− 8x + 25
dx
)
2
)
dx
(
(
)
=
∫⎡
dx
(
)
⎤
⎣ x − 8x + 16 + 25 − 16 ⎦
2
)
(
u=3tanθ→du=3tanθsecθdθ
2
=
∫⎡
dx
(
)
(
)
2
2
=
∫
du
=
2
2
⎤
x − 8x + 25
u +9
x−4 +9
⎢⎣
⎥⎦
u=x−4→du=dx
3sec θ tanθ
3sec θ tanθ
3 sec θ tanθ
1 tanθ
dθ = ∫
dθ =
dθ =
dθ =
∫
∫
2
2
2
81 sec 2 θ
27 ∫ sec 3 θ
⎡9 tan2 θ + 1 ⎤
9 tan2 θ + 9
⎣
⎦
2
2
(
)
)
1 sinθ
1
1 1
1
⋅ cos3 θdθ =
cos2 θsinθdθ = −
⋅ cos3 θ + C = − cos3 θ + C =
∫
∫
27 cosθ
27
27 3
81
w=cosθ→−dw=sinθdθ
3
−
1 ⎛
3 ⎞
1
⋅⎜
+C= −
⎟
2
81 ⎝ u + 9 ⎠
3 u2 + 9
(
)
3/2
+C= −
1
(
)
3 ⎡ x − 4 + 9⎤
⎢⎣
⎥⎦
u2+9
θ
3
2
3/2
+C= −
u
(
1
3 x − 8x + 25
2
)
3/2
+C
12)
∫ x sin ( x )dx
2
∫ x sin ( x )dx =
( )
1
sinudu
2∫
2
1
u=x 2 →du=2xdx→ du=xdx
2
13)
1
1
= − cosu + C = − cos x 2 + C
2
2
∫ x sin2xdx
∫ x sin2x =
uv − ∫ v du
u=x dv=sin2xdx
1
du=dx v= sin2x dx=− cos 2x
2
1
1
1
1
= − x cos 2x − ∫ − cos 2x dx = − x cos 2x + ∫ cos 2x dx =
2
2
2
2
∫
1
1 1
1
1
− x cos 2x + ⋅ sin2x + C = − x cos 2x + sin2x + C
2
2 2
2
4
14)
∫x
∫x
2
2
sin2xdx
uv − ∫ v du
sin2xdx =
u=x 2
du=2xdx
dv=sin2xdx
1
v=− cos 2x
2
1
1
1
= − x 2 cos 2x − ∫ − cos 2x ⋅ 2x dx = − x 2 cos 2x + ∫ x cos 2x dx =
2
2
2
1
1
1
1
− x 2 cos 2x + uv − ∫ v du = − x 2 cos 2x + x sin2x − ∫ sin2x dx =
2
2
2
2
u=x
dv=cos 2xdx
du=dx
1
v= sin2x
2
1
1
1
1
1
1 1
− x 2 cos 2x + x sin2x − ∫ sin2x dx = − x 2 cos 2x + x sin2x − ⋅ − cos 2x + C =
2
2
2
2
2
2 2
1
1
1
− x 2 cos 2x + x sin2x + cos 2x + C
2
2
4
(
15)
x +1
dx
2
+4
∫x
x +1
dx =
2
+4
∫x
(
1
ln x 2 + 4
2
∫x
)
abs.value is not needed
because x 2 +4>0 for all x
16)
)
2
+
x
1
dx + ∫ 2
dx =
+4
x +4
1
x
tan−1 + C
2
2
x +1
dx Use partial fractions
2
−4
∫x
1 1
1
1
1
x
du + ∫ 2
dx = ln u + tan−1 + C =
∫
2 u
2
2
2
x +4
1
1
1
−1 x
u=x 2 +4→ du=xdx
dx=
∫ x 2 +a2 a tan a +C
2
x +1
x +1
=
2
x −4
x+2 x−2
(
)(
) (
x +1
A
B
=
+
⇒ x + 1⇒ A x − 2 + B x + 2
x+2 x−2
x+2 x−2
)(
(
)
) (
)
3
1
x = −2 ⇒ −1= −4A ⇒ A =
4
4
⎛ 1/ 4 3 / 4 ⎞
x +1
1
1
3
1
1
3
∫ x 2 − 4 dx = ∫ ⎜⎝ x + 2 + x − 2 ⎟⎠ dx = 4 ∫ x + 2 dx + 4 ∫ x − 2 dx = 4 ln x + 2 + 4 ln x − 2 + C
x = 2 ⇒ 3 = 4B ⇒ B =
17)
∫x
x +1
dx Use partial fractions-repeated linear factors
− 4x + 4
2
∫x
2
x +1
dx =
− 4x + 4
x +1
( x − 2)
2
x +1
dx =
− 4x + 4
A
B
+
x−2 x−2
=
2
∫x
(
)
2
(
(
18)
∫x
(
(
x +1
x−2
x +1
x x−2
)
)
−2
=
(
7
1
⇒B = −
4
4
3
x−2
2∫
(
19)
)
−2
x +1
∫ x (x
2
2
(
)
+1
x +1
)
x x +1
2
2
dx
)
⎛
x +1
1
3
∫ x 2 − 4x + 4 dx = ∫ ⎜⎜ x − 2 + x − 2
⎝
(
dx = ln x − 2 + 3 ⋅
A
B
C
+
+
x x−2 x−2
x = 2 ⇒ 3 = 2C ⇒ C =
2 = −B +
( x − 2)
2
1
x−2
−1
(
)
−1
+ C = ln x − 2 −
)
⎞
⎟ dx =
2
⎟⎠
3
+C
x−2
dx Use partial fractions-repeated linear factors
2
2
)
x +1
⇒ x + 1= A x − 2 + B x = 2 ⇒ B = 3 x = 0 ⇒ 1= −2A + B ⇒
1= −2A + 3 ⇒ −2 = −2A ⇒ A = 1
1
∫ x − 2 dx + 3 ∫ x − 2
x +1
∫ ( x − 2) ( x − 2) dx = ∫
dx =
3
2
)
(
)
2
(
)
x + 1= A x − 2 + Bx x − 2 + Cx
2
1
1
3
x = 1⇒ 2 = A − B + C ⇒ 2 = − B + ⇒
4
4
2
⎛
⎞
x +1
1/ 4 1/ 4
3/2
1 1
1
1
⎜
dx
=
−
+
∫ x x−2 2
∫ ⎜ x x − 2 x − 2 2 ⎟⎟ dx = 4 ∫ x dx − 4 ∫ x − 2 dx +
⎝
⎠
x = 0 ⇒ 1= 4A ⇒ A =
(
)
(
1
1
3 1
ln x − ln x − 2 + ⋅
x−2
4
4
2 −1
(
)
−1
)
1
1
3
+ C = ln x − ln x − 2 −
+C
4
4
2 x−2
(
)
dx Use partial fractions-repeated linear and non-repeated quadratic factors
=
A B Cx + D
+ 2+ 2
⇒ x + 1= Ax x 2 + 1 + B x 2 + 1 + Cx + D x 2 + 1
x x
x +1
(
) (
We will find A,B, C, and D by equating coefficients.
) (
)(
)
(
) (
) (
)(
)
If x + 1= Ax x 2 + 1 + B x 2 + 1 + Cx + D x 2 + 1 , then coefficients of the x 3 ,x 2 ,x and constant terms on
both sides must be equal. Since no x 3 or x 2 appear on the left side, their coefficients on the left side
are zero.
(
) (
) (
)
x + 1= Ax x 2 + 1 + B x 2 + 1 + Cx + D x 2 ⇒ x + 1= Ax 3 + Ax + Bx 2 + B + Cx 3 + Cx + Dx 2 ⇒
(
)
(
)
x + 1= A + C x 3 + B + D x 2 + Ax + B
Since the coefficients of both the x 3 and x 2 terms on the left side are zero, we have
A + C = 0 ⇒ C = −A and B + D = 0 ⇒ D = −B . This leaves
x + 1= Ax + B ⇒ A = 1,B = 1,C = −1, and D = −1⇒ ∫
1
∫ x dx + ∫ x
−2
dx − ∫
⎛ 1 1 −x − 1⎞
dx
=
∫
⎜⎝ x + x 2 + x 2 + 1 ⎟⎠ dx =
x2 x2 + 1
x +1
(
(
)
)
x
1
1
dx − ∫ 2
dx = ln x − x −1 − ln x 2 + 1 − tan−1 x + C =
2
x +1
x +1
2
1
u=x +1→ du=xdx
2
2
ln x −
20)
(
)
1 1
− ln x 2 + 1 − tan−1 x + C
x 2
x3
∫ 4 + x 2 dx
(
)
3
2tanθ
x3
8 tan3 θ
16 tan3 θsec 2 θ
2
2
dx
=
⋅
2sec
θdθ
=
⋅
2sec
θdθ
=
∫ 4 + x2
∫ 4 + 2tanθ 2
∫ 4 + 4 tan2 θ
∫ 4 1+ tan2 θ dθ =
(
(
)
2
)
x=2tanθ→dx=2sec θdθ
4∫
tan θsec θ
dθ = 4 ∫ tan3θdθ = 4 ∫ tan2 θ tanθdθ = 4 ∫ sec 2 θ − 1 tanθ =
sec 2 θ
3
(
2
4 ∫ sec θ tanθdθ − 4 ∫ tanθdθ =
2
tan2 θ
4⋅
2 2
u=tanθ→du=sec θdθ
2
(
)
(
)
2
⎛ x⎞
− 4ln sec θ + C = 2 ⎜ ⎟ − 4ln
⎝ 2⎠
)
x
1
x2
2
− 4 ⋅ ln x + 4 + 4ln2 + C1 =
− 2ln x 2 + 4 + C, where C = C1 − 4ln2
2
2
2
x2+4
θ
2
x
x2 + 4
+C=
2
x
x3
as x 2 ⋅
and use the direct substitution
2
4 + x2
4+x
1
x3
1 u− 4
1 ⎛ 4⎞
u = 4 + x 2 . Then du = xdx and x 2 = u − 4 ⇒ ∫
dx = ∫
du = ∫ ⎜ 1− ⎟ du
2
2
2
u
2 ⎝ u⎠
4+x
Alternate solution: Write
∫
21)
∫
8∫
x3
4 + x2
x3
4 + x2
dx
( 2tanθ)
3
dx =
∫
4 + 4 tan2 θ
⋅ 2sec 2 θdθ =
2
x=2tanθ→dx=2sec θdθ
∫
8 tan3 θ
(
4 1+ tan2 θ
)
⋅ 2sec 2 θdθ =
16 tan3 θsec 2 θ
∫ 4sec 2 θ dθ =
tan3 θsec 2 θ
dθ = 8 ∫ tan3 θsec θdθ = 8 ∫ tan2 θsec θ tanθdθ = 8 ∫ sec 2 θ − 1 sec θ tanθdθ =
sec θ
(
)
3
⎛ x2 + 4 ⎞
u3
8
8 ⎛ x2 + 4 ⎞
8∫
u2 − 1 du = 8 ⋅ − 8u + C = sec 3 θ − 8sec θ + C = ⋅ ⎜
⎟ − 8⎜
⎟ +C=
3
3
3 ⎝
2 ⎠
2 ⎠
⎝
u=sec θ→du=sec θ tanθdθ
(
(
)
)
2
2
⎛ x2 + 4
⎞ 1 2
8 x +4 x +4
⋅
− 4 x2 + 4 + C = x2 + 4 ⋅ ⎜
− 4⎟ =
x + 4 x2 − 8 + C
3
⎝ 3
⎠ 3
8
(
x2+4
)
x
θ
2
x
x3
Alternate Solution: As in the previous problem, Write
as x 2 ⋅
and use the direct substitution
2
4 + x2
4+x
1
x3
1 u− 4
1 ⎛
4 ⎞
u = 4 + x 2 . Then du = xdx and x 2 = u − 4 ⇒ ∫
dx = ∫
du = ∫ ⎜ u −
⎟ du =
2
2
2 ⎝
u
u⎠
4 + x2
1
u1/2 − 4u−1/2 du
∫
2
(
)
22)
x3
∫
x −1
2
x3
∫
x2 − 1
∫ sec
4
dx
dx =
sec 3 θ
∫
sec 2 θ − 1
⋅ sec θ tanθdθ =
∫
sec 3 θ
tan2 θ
x=sec θ→dx=sec θ tanθdθ
θdθ = ∫ sec 2θsec 2 θdθ =
tan θ
tanθ +
+ C = x2 − 1 +
3
2
3
∫ (1+ tan θ) sec
2
(
) +C=
=tanθ→du=sec θdθ
sec 3 θ
∫ tanθ sec θ tanθdθ =
θdθ = ∫ sec 2 θdθ + ∫ tan2 θsec 2 θdθ =
3
x2 − 1
3
2
⋅ sec θ tanθdθ =
( x − 1)
−1+
3/2
2
x2
3
+C
x
x2-1
θ
1
Or, use a solution similar to problems 20 and 21.
x3
dx
23) ∫ 2
x −1
x3
sec 3 θ
sec 3 θ
sec 3 θ
dx
=
⋅
sec
θ
tanθdθ
=
⋅
sec
θ
tanθdθ
=
∫ x 2 − 1 ∫ sec 2 θ − 1
∫ tan2 θ
∫ tanθ sec θdθ =
x=sec θ→dx=sec θ tanθdθ
sec θ
sec θ
2
∫ tanθ dθ = ∫ tanθ ⋅ sec θdθ =
4
2
(1+ tan θ) sec
2
∫
2
tanθ
θdθ =
1+ u2
1
∫ u du = ∫ u du + ∫ udu =
2
u=tanθ→du=sec θdθ
u
tan θ
x −1
1
x2
+ C = ln tanθ +
+ C = ln x 2 − 1 +
+ C1 = ln x 2 − 1 +
+ C, where
2
2
2
2
2
1
C = C1 −
2
Use the same triangle as in #22. Also, the problem can be solved as in #20 and 21 with u-substitution.
ln u +
2
2
2
(
)
24)
1
∫
3 − 5x 2
dx
1
∫
3 − 5x
1
2
dx =
3 cosθ
∫
5
3 cosθ
1
∫
3−
dθ =
( 5x )
1
2
∫ 1dθ =
5
dx =
1
1
∫
5
3−u
u= 5x→
1
5
1
5
du =
1
∫
5
5
3 cosθ
3 − 3sin θ
2
dθ =
1
∫
5
(
sin−1
u
3
+C=
1
5
sin−1
5x
3
+C=
3 cosθ
3 1− sin2 θ
u= 3 sinθ→du= 3 cosθdθ
dx
1
θ+C =
2
1
5
sin−1
)
15x
+C
3
3
u
θ
3-u2
25)
∫ (csc
)
x − 2csc 2 x cot 2 x + cot 4 x dx
4
∫ (csc x − 2csc x cot
∫ 1 dx = ∫ 1dx = x + C
4
2
2
)
x + cot 4 x dx =
∫ (csc
2
)
2
x − cot 2 x dx =
∫ (1+ cot
2
)
2
x − cot 2 x dx =
2
π/2
26)
∫ csc
3
x cot x
π/6
π/2
∫ csc
π/2
3
x cot x =
π/6
∫ csc
π/6
1
2
x csc x cot x dx =
− ∫ u du
2
∫ x ln ( x + 2) dx
u3
8 1 7
= ∫ u du =
= − =
3 1 3 3 3
1
2
2
b
u=csc x→du=csc x cot xdx→−du=csc x cot x a
− f t dt = f t dt
x=π/6→u=csc π/6 →u=2
x=π/2→u=csc π/2 →u=1
b
a
( )
( )
27)
2
2
∫ () ∫ ()
dθ =
(
)
x2
∫ x ln x + 2 dx = uv − ∫ v du =
(
)
u=ln x+2
du=
=
x2
2
(
)
ln x + 2 −
1
2
1
x+2
2
dv=xdx
dx
v=
x
∫x
2
)
x2
1
⋅
2 x+2
dx =
x2
2
(
)
ln x + 2 −
1
x2
2∫x+2
dx =
x2
2
(
)
ln x + 2 −
1 ⎛
4 ⎞
x −2+
dx
2 ∫ ⎜⎝
x + 2 ⎟⎠
↑
long division
2
2
∫ x dx + ∫ 1dx − 2 ∫
1
x+2
dx =
x2
2
(
)
ln x + 2 −
1 x2
⋅
+ x − 2ln x + 2 + C
2 2
(
)(
)
x+2 x−2
x2
x2 − 4 + 4 x2 − 4
4
4
4
=
=
+
=
+
= x − 2+
x+2
x+2
x+2 x+2
x+2
x+2
x+2
Alternate to long division:
28)
(
ln x + 2 − ∫
sin3xdx
1
du = 2xdx and v = ∫ sin3x dx = − cos3x + C ⇒ ∫ x 2 sin3x dx = uv − ∫ v du =
3
1
1
1
2
− x 2 cos3x − ∫ − cos3x ⋅ 2xdx = − x 2 cos3x + uv − ∫ v du =
3
3
3
3
(
)
u=x dv=cos3x
1
du=dx v= sin3x
3
⎞ 2 1
1
2⎛ 1
1
2
2
− x 2 cos3x + ⎜ x sin3x ⎟ − ∫ sin3x + C = − x 2 cos3x + x sin3x − ∫ sin3x dx =
3
3⎝3
3
9
9
⎠ 3 3
⎞
1
2
2⎛ 1
1
2
2
− x 2 cos3x + x sin3x − ⎜ − cos3x ⎟ + C = − x 2 cos3x + x sin3x +
cos3x + C
3
9
9⎝ 3
3
9
27
⎠
29)
∫ sec x tan xdx
∫ sec x tan xdx = ∫ sec
3
5
3
=
5
(
)
2
(
(
2
)
2
2
2
4
2
∫ u u − 1 du = ∫ u u − 2u + 1 du =
u=sec x→du=sec x tan xdx
)
(
2
∫(
)
u 6 − 2u 4 + u 2 du = =
u 7 2u 5 u 3
−
+
+C =
7
5
3
sec 7 x sec 5 x sec 3 x
−
+
+C
7
5
3
30)
∫ sin
2
∫ sin
t cos2 t dt
2
t cos2 t dt =
∫ 2 (1− cos 2t ) 2 (1+ cos 2t ) dt = 4 ∫ (1− cos 2t ) dt = 4 ∫ sin
1
1
1
2
1
2
2t dt =
1 1
1
1
1
1
1 1
1− cos 4t dt = ∫ 1− cos 4t dt = ∫ 1dt − ∫ cos 4t dt = t − ⋅ sin4t + C =
∫
4 2
8
8
8
8
8 4
(
)
1
sin2 x= 1−cos 2x with x=2t
2
(
)
1
1
t−
sin4t + C
8
32
)
2
x tan4 x sec x tan x dx = ∫ sec 2 x tan2 x dx = ∫ sec 2 x sec 2 x − 1 sec x tan x
(
)
1/2
∫ tan ( 2x ) dx
Use Integration by parts with u = tan−1 2x and dx = dx
−1
31)
0
u = tan−1 2x ⇒ du =
1
( )
1+ 2x
2
( )
⋅ 2x ′ dx =
2
1/2
1/2
1+ 4x
dx and dv = dx ⇒ v = x ⇒ uv 0 −
2
∫ udv =
0
CHAIN RULE
x tan
1
2
32)
∫
−1
( 2x )
∫
()
9 − x2
x3
9 − x2
27 ∫
cos θ
(
∫ 1+ 4x
= x tan
2
−1
( 2x )
0
1/2
0
1
2
8∫w
−
( 2x )
1/2
0
1
2
w=1+4 x → dw=dx
8
x=0→w=1 x=1/2→w=2
()
−
1
4
2
=
ln w
1
()
∫
27 sin3 θ
⋅ 3 cos θdθ = 27 ⋅ 3 ∫
9 − 9 sin2 θ
sin3 θ cos θ
(
9 1− sin2 θ
)
dθ =
(
27 ⋅ 3 sin3 θ cos θ
∫
3
cos 2 θ
)
dθ = 27 ∫ sin3 θ dθ = 27 ∫ sin2θ sinθdθ = 27 ∫ 1− cos 2 θ sinθ dθ =
)( −du)
u=cos θ→du=− sin θdθ=− du=sin θdθ
(
)
= 27 ∫ u − 1 du = 27 ⋅
2
(
u3
3
− 27u + C = 9u3 − 27u + C =
)
9 cos θ − 27 cos θ + C = 9 cos θ cos θ − 3 + C = 9 ⋅
3
⎛ 9 − x 2 − 27 ⎞
3 9−x ⎜
⎟⎠ + C =
9
⎝
2
= x tan
du
−1
⎛1
1 ⎞ 1 π 1
π 1
tan−1 0 − ⎜ ln 2 − ln1⎟ = ⋅ − ln 2 = − ln 2 ⎡⎣ tan−1 0 = 0 and ln1 = 0 ⎤⎦
⎝4
2
4 ⎠ 2 4 4
8 4
1
sin3 θ cos θ
27 ∫ 1− u
2x
dx
dx =
2
−
0
tan−1 1 −
x3
1/2
1/2
2
9 − x2
3
(
⋅ −18 − x
2
9 − x2 ⎛ 9 − x2
3
)+C = −
⎜⎝
9
(
⎞
− 3⎟ + C =
⎠
9 − x 2 x 2 + 18
3
)+C
dθ =
3
x
θ
9-x2
33)
∫t
dt
[HINT: Use completing the square]
− 6t + 13
2
tdt
∫
t − 6t + 13
2
u
∫
u +4
2
1
w
2∫
−1/2
∫
=
u2 + 4 + ln
2
u +4
2
(
)
1
2∫
w
=
tdt
∫
( t − 3)
+∫
dw
1
2
w=u +4→dw=2udu→ dw=du
2
2 sec θ
2
dθ =
1
⋅
1
2 1/ 2
w1/2 +
2
+4
u+3
=∫
(
)
2
+ 4 ⎤ + ln 2 + C =
⎦
du =
u2 + 4
u=t−3→t=u+3→du=dt
3
4 tan2 θ + 4
⋅ sec θ tan θdθ =
u=2 tan θ→du=2 tan θ sec θ
3
tanθ dθ =
2∫
u2 + 4 +
3
2
+ C = u2 + 4 + ln 2 − ln u2 + 4 = u2 + 4 + ln 2 −
u +4
2
1
t − 3 + 4 − ln ⎡ t − 3
2 ⎣
C1 = C + ln 2
2
1
du =
sec θ tanθ
4
)
− 6t + 9 + 13 − 9
3
du + ∫
dw + 3 ∫
(t
tdt
t 2 − 6t + 13 −
1
2
(
ln sec θ + C =
1
2
(
)
ln u2 + 4 + C =
)
ln t 2 − 6t + 13 + C1, where
4+u2
u
θ
2
34)
∫
(
(
x 2 + x + 16
)(
x +1 x − 3
)
x 2 + x − 16
)(
x +1 x − 3
)
=
2
Use the method of partial fractions-repeated linear factors.
2
A
x +1
+
B
x−3
(
x = −1⇒ 1− 1− 16 = A −1− 3
+
)
2
C
(
x−3
)
2
(
⇒ x 2 + x − 16 = A x − 3
)
2
(
)(
)
(
)
+B x − 3 x +1 + C x +1 ⇒
⇒ −16 = 16A ⇒ A = −1 x = 3 ⇒ 9 + 3 − 16 = 2C ⇒ −4 = 2C ⇒ C = −2
To find B, set x = 0 along with the values of A and C.
( )
−16 = A −3
∫
2
( )( )
()
+ B −3 1 + C 1 ⇒ −16 = −9 − 3B − 1⇒ −16 = −10 − 3B ⇒ −6 = 3B ⇒ B = −2
x 2 + x − 16
( x + 1) ( x − 3)
2
=
−1
2
∫ x + 1 dx + ∫ x − 3 dx + ∫
1
− ln x + 1 + 2ln x − 3 −
x−3
−1
(
35)
∫x
3
3 − x 2 dx
)
−1
−1
( x − 3)
2
dx = − ∫
1
x +1
dx + 2 ∫
1
x−3
(
)
2
dx − ∫ x − 3 dx =
(
)
2
x−3
1
1
+ C = − ln x + 1 + 2ln x − 3 +
+ C = ln
+
+C
x−3
x +1
x−3
3
2
∫ x 3 − x dx =
∫(
3 sinθ
∫
3 ∫ (1− cos θ ) ( ) cos
)
3
(
)
3 − 3 sin2 θ ⋅ 3 cos θdθ = ∫ 3 3 ⋅ sin3 θ 3 1− sin2 θ ⋅ 3 cos θdθ =
∫
− ∫ 9 3 (1− u ) u
∫
3 3 3sin3 θ cos2 θ cosθdθ = 9 3 sin3 θcos2θ = 9 3 sin2 θcos2 θsinθdθ =
9
2
2
θsinθ =
2
2
(
)
(
)
du = 9 3 ∫ u2 − 1 u2du = 9 3 ∫ u4 − u2 du =
u=cosθ→du=− sinθdθ→−du=sinθdθ
u5
u3
9 3
3 3 cos3 θ
−9 3⋅ +C=
cos5 θ − 3 3 cos3 θ + C =
3cos2 θ − 5 + C =
5
3
5
5
3/2
3
2
⎡
⎤
⎡ 3 3 − x2
⎤
3 − x2
3 3 ⎛ 3 − x2 ⎞ ⎢ ⎛ 3 − x2 ⎞
3
3
⎥
⎢
⎥+C=
⋅⎜
3
−
5
+
C
=
⋅
−
5
⎟
⎜
⎟
⎥
5 ⎝
5
⎢
⎥
3
3 ⎠ ⎢⎢ ⎝
3 ⎠
3 3 ⎣
⎦
⎥⎦
⎣
(
9 3⋅
(
(3 − x ) ⋅ (3 − x
5
2
3/2
(3 − x ) ( −x
− 5) + C =
2
2
3/2
5
2
−2
)
(
)
) + C = − 1 (3 − x ) ( x
5
2
)
3/2
2
)
+2 +C
3
x
θ
3-x2
36)
x3 − 8
∫ x + 2 dx The integrand is an improper rational function. Therefore, begin by performing long-division
on the integrand.
x 2 − 2x + 4
x + 2 x 3 + 0x 2 + 0x − 8
(
− x 3 + 2x 2
)
_____________
(
− 2x 2
− −2x 2 − 4x
)
−8
_____________
4x − 8
4x + 8
_____________
− 16
⎛ 2
x3 − 8
16 ⎞
x3
dx
=
x
−
2x
+
4
−
dx
=
− x 2 + 4x − 16ln x + 2 + C
∫ x+2
∫ ⎜⎝
x + 2 ⎟⎠
3
)(
(
)
x + 2 x 2 − 2x + 4
x 3 − 8 x 3 + 8 − 16 x 3 + 8
16
16
16
=
=
−
=
−
= x 2 − 2x + 4 −
NOTE:
x+2
x+2
x+2 x+2
x+2
x+2
x+2
(
)
a3 +b3 = a+b ⎛⎜⎝ a2 −ab+b2 ⎞⎟⎠
37)
∫ (x
(x
20
2
)(
20
2
dx Use partial fractions-non-repeated linear and quadratic factors
)
+ 4 x −1
)(
)
+ 4 x −1
=
Ax + B
C
+
⇒ 20 = Ax + B x − 1 + C x 2 + 4 . To find A, B, and C, first set x = 1 in the
2
x + 4 x −1
(
)(
)
(
)
last equation.
(
)(
)
(
)
x = 1: 20 = Ax + B x − 1 + C x 2 + 4 ⇒ 20 = 5C ⇒ C = 4 . Setting x = 0 and C = 4 allows for finding B.
( )( ) ( )
x = 0 : 20 = B −1 + 4 4 ⇒ 20 = −B + 16 ⇒ B = −4 . Now try x = -1 with B = -4 and C = 4 to find A.
(
)( ) ( )
x = −1: 20 = −A − 4 −2 + 4 5 ⇒ 20 = 2A + 8 + 20 ⇒ 2A = −8 ⇒ A = −4
∫ (x
⎛ −4x − 4
4 ⎞
2x
4
4
dx = ∫ ⎜ 2
+
dx = −2 ∫ 2
dx − ∫ 2
dx + ∫
dx =
⎟
x −1
⎝ x + 4 x − 1⎠
x +4
x +4
+ 4 x −1
20
2
(
)(
)
)
−2ln x 2 + 4 −
u=x 2 +4→du=2xdx
1 1
x
⋅ tan−1 + 4ln x − 1 + C
4 2
2 u=x−1→du=dx
1
x
tan−1 +C
a
a
π/4
38)
∫ sin
5
2x cos3 2x
0
π/4
∫
π/4
∫
sin5 2x cos3 2xdx =
0
0
1
(
)
1
u=sin2x→ du=cos 2xdx
2
π
π
x=0→u=sin0=0 x= →u=sin =1
4
2
∫e
−2x
∫ sin
5
(
)
2x 1− sin2 2x cos 2xdx =
0
1
1
1
1 ⎛ u6 u8 ⎞
1 ⎛ 1 1⎞ 1 1
1
5
7
= ∫ u − u du = ⎜ − ⎟ = ⎜ − ⎟ = ⋅
=
20
2⎝ 6 8 ⎠
2 ⎝ 6 8 ⎠ 2 24 48
0
(
1 5
u 1− u2 du
∫
20
39)
π/4
sin5 2x cos2 2x cos 2xdx =
)
sin5x dx
dv = e −2x dx
u = sin5x
1
v = ∫ e −2x dx = − e −2x
2
du = 5cos5xdx
1
5
1
5
sin5x dx = uv − ∫ v du = − e −2 x sin5x − ∫ − e −2 x cos 5x dx = − e −2 x sin5x +
e −2x cos5x dx =
2
2
2
2
⎤
1
5⎡
1
5⎡ 1
5 −2x
− e −2 x sin5x +
uv − v du ⎤⎦ = − e −2 x sin5x + ⎢ − e −2x cos5x −
e sin5x dx ⎥ ⇒
⎣
2
2
2
2⎣ 2
2
⎦
−2 x
∫e
∫
−2 x
∫
u=cos5x
∫
dv=e
dx
1
du=−5sin5xdx v=− e−2 x
2
1
5
25 −2 x
sin5x dx = − e −2 x sin5x − e −2 x cos 5x −
e sin5x dx ⇒
2
4
4 ∫
25 −2 x
1 −2 x
5 −2 x
−2 x
∫ e sin5x dx + 4 ∫ e sin5x dx = − 2 e sin5x − 4 e cos 5x + C ⇒
29 −2 x
1
5
e sin5x dx = − e −2 x sin5x − e −2 x cos 5x + C ⇒
∫
4
2
4
∫e
−2 x
∫e
−2 x
2
40)
sin5x dx =
1
∫
4x − x 2
1
⎤
4 ⎡ 1 −2x
5
4
− e sin5x − e −2x cos5x ⎥ + C1, where C1 =
C
⎢
29 ⎣ 2
4
29
⎦
dx First, use completing the square on the integrand.
(
)
(
4x − x = − x − 4x + 4 + 4 = 4 − x − 2
2
2
2
) ∫
2
1
0
∫
− π/6
2 cos θ
(
0
4 1− sin θ
2
)
dθ =
∫
− π/6
2 cos θ
4 cos θ
2
0
dθ =
∫
− π/6
2
1
4x − x
2 cos θ
2 cos θ
2
dx =
1
0
dθ =
∫
(
4− x−2
∫ 1dθ = θ
− π/6
0
1
0
− π/6
)
2
dx =
∫
−1
0
1
4−u
2
du =
u= x −2→du=dx
x =1→u=−1 x =0→x =−2
⎛ π⎞ π
= 0−⎜− ⎟ =
⎝ 6⎠ 6
∫
− π/6
1
4 − 4 sin θ
2
u=2 sin θ→du=2 cos θdθ
1
π
u=−1→− =sin θ→θ=−
2
6
u=0→0=sin θ→θ=0
⋅ 2 cos θdθ =
41)
∫x
∫
dx
2
dx
x
2
Use the direct substitution u= 4x, allowing for the trig substitution u = 3sec θ
16x 2 − 9
=
16x − 9
2
∫
1
x
( 4x )
2
dx =
2
−9
1
∫
4 u
du
=4
2
∫
u −9
2
16
du
u
2
=4
u −9
2
∫
3 sec θ tan θ
dθ =
9 sec θ 9 sec θ − 9
2
2
u= 3 sec θ→du= 3 sec θ tan θdθ
1
u= 4 x→ du= dx
4
u
4
3 sec θ tan θ
4
∫
9
4
9
∫
(
)
sec θ 9 sec θ − 1
2
cos θ dθ =
2
4
9
dθ =
sin θ + C =
4
9
= x→
u
2
3 sec θ tan θ
4
∫
9
sec θ ⋅ 9 tan θ
2
u −9
2
2
⋅
2
=x
16
u
+C=
4
9
dθ =
4
3 sec θ tan θ
∫
9 sec
16x − 9
2
θ ⋅ 3 tan θ
2
⋅
4x
dθ =
16x − 9
4
1
dθ =
∫
9 sec θ
2
+C=
9x
u
+C
u2-9
θ
3
42)
∫ cos (ln x ) dx
∫ cos (ln x ) dx =
( )
Begin with integration by parts with u = cos ln x
∫
uv − v du
( )
sin( ln x )
d
du=− sin( ln x ) [ ln x ]=−
u=cos ln x
dx
⇑
CHAIN RULE!
( )
x cos ln x +
∫
( )
cos( ln x )
d
du=cos( ln x ) [ ln x ]=
dx
We now have
( )
x
dv =dx
( ) x dx =x cos (ln x ) +
sin ln x
x
( ) ∫
= x cos ln x + x sin ln x −
dv =dx
C1
2
x
v=x
( )
∫
( )
2 cos ln x dx = x cos ln x + x sin ln x + C1 ⇒ cos ln x dx =
where C =
( ) x dx =x cos (ln x ) + x sin (ln x ) −
cos ln x
∫ cos (ln x ) dx = x cos (ln x ) + x sin (ln x ) − ∫ cos (ln x ) dx ⇒
( )
∫ sin (ln x ) dx =
v=x
( )
uv − v du
u=sin ln x
∫
x
( ) ∫
= x cos ln x − −
( )
( ) + C,
x cos ln x + x sin ln x
2
∫ cos (ln x ) dx