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317
CHAPTER 8
Principles of Integral Evaluation
EXERCISE SET 8.
- u = 3 − 2 x, du = − 2 dx, −
u^3 du = −
u^4 + C = −
(3 − 2 x)^4 + C
- u = 4 + 9x, du = 9dx,
u^1 /^2 du =
u^3 /^2 + C =
( 4 + 9x)^3 /^2 + C
- u = x^2 , du = 2xdx,
sec^2 u du =
tan u + C =
tan(x^2 ) + C
- u = x^2 , du = 2xdx, 2
tan u du = −2 ln | cos u | + C = −2 ln | cos(x^2 )| + C
- u = 2 + cos 3x, du = −3 sin 3xdx, −
du u
ln |u| + C = −
ln(2 + cos 3x) + C
- u = 3 x 2 , du =
dx,
du 4 + 4u^2
du 1 + u^2
tan−^1 u + C =
tan−^1 (3x/2) + C
- u = ex, du = exdx,
sinh u du = cosh u + C = cosh ex^ + C
- u = ln x, du =
x dx,
sec u tan u du = sec u + C = sec(ln x) + C
- u = cot x, du = − csc^2 xdx, −
eu^ du = −eu^ + C = −ecot^ x^ + C
- u = x^2 , du = 2xdx, 1 2
√du 1 − u^2
=^1
sin−^1 u + C =^1 2 sin−^1 (x^2 ) + C
- u = cos 7x, du = −7 sin 7xdx, −
u^5 du = −
u^6 + C = −
cos^6 7 x + C
- u = sin x, du = cos x dx,
du u
u^2 + 1
= − ln
1 + u^2 u
∣ +^ C^ =^ −^ ln
1 + sin^2 x sin x
∣ +^ C
- u = ex, du = exdx,
√du 4 + u^2
= ln
u +
u^2 + 4
ex^ +
e^2 x^ + 4
+ C
- u = tan−^1 x, du =
1 + x^2 dx,
eu^ du = eu^ + C = etan − (^1) x
- u =
x − 2 , du =
x − 2
dx, 2
eu^ du = 2eu^ + C = 2e
√x− 2
- u = 3x^2 + 2x, du = ( 6x + 2)dx,
cot u du =
ln | sin u| + C =
ln sin | 3 x^2 + 2x| + C
- u =
x, du =
x dx,
2 cosh u du = 2 sinh u + C = 2 sinh
x + C
318 Chapter 8
- u = ln x, du = dx x ,
du u = ln^ |u|^ +^ C^ = ln^ |ln^ x|^ +^ C
- u =
x, du =
x dx,
2 du 3 u^
e−u^ ln 3^ du = −
ln 3 e−u^ ln 3^ + C = −
ln 3
√x
- u = sin θ, du = cos θdθ,
sec u tan u du = sec u + C = sec(sin θ) + C
- u =
x , du = −
x^2 dx, −
csch^2 u du =
coth u + C =
coth
x
+ C
∫ (^) dx √ x^2 − 3
= ln
∣x^ +^
x^2 − 3
∣ +^ C
- u = e−x, du = −e−xdx, −
du 4 − u^2
ln
∣∣^ 2 +^ u 2 − u
∣∣ + C = − 1
ln
∣∣^ 2 +^ e
−x 2 − e−x
∣∣ + C
- u = ln x, du =
x dx,
cos u du = sin u + C = sin(ln x) + C
- u = ex, du = exdx,
√ex^ dx 1 − e^2 x^
√du 1 − u^2
= sin−^1 u + C = sin−^1 ex^ + C
- u = x−^1 /^2 , du = −
2 x^3 /^2 dx, −
2 sinh u du = −2 cosh u + C = −2 cosh(x−^1 /^2 ) + C
- u = x^2 , du = 2xdx, 1 2
du sec u
=^1
cos u du =^1 2 sin u + C =^1 2 sin(x^2 ) + C
- 2 u = ex, 2 du = exdx,
√^2 du 4 − 4 u^2
= sin−^1 u + C = sin−^1 (ex/2) + C
- 4 −x 2 = e−x (^2) ln 4 , u = −x^2 ln 4, du = − 2 x ln 4 dx = −x ln 16 dx,
−
ln 16
eu^ du = −
ln 16 eu^ + C = −
ln 16 e−x (^2) ln 4
ln 16 4 −x 2
- 2 πx^ = eπx^ ln 2,
2 πx^ dx =
π ln 2 eπx^ ln 2^ + C =
π ln 2 2 πx^ + C
EXERCISE SET 8.
- u = x, dv = e−xdx, du = dx, v = −e−x;
xe−xdx = −xe−x^ +
e−xdx = −xe−x^ − e−x^ + C
- u = x, dv = e^3 xdx, du = dx, v =
e^3 x;
xe^3 xdx =
xe^3 x^ −
e^3 xdx =
xe^3 x^ −
e^3 x^ + C
- u = x^2 , dv = exdx, du = 2x dx, v = ex;
x^2 exdx = x^2 ex^ − 2
xexdx.
For
xexdx use u = x, dv = exdx, du = dx, v = ex^ to get ∫ xexdx = xex^ − ex^ + C 1 so
x^2 exdx = x^2 ex^ − 2 xex^ + 2ex^ + C
320 Chapter 8
- u = ln(2x + 3), dv = dx, du = 2 2 x + 3 dx, v = x;
ln(2x + 3)dx = x ln(2x + 3) −
2 x 2 x + 3 dx
but
2 x 2 x + 3 dx =
2 x + 3
dx = x −
ln(2x + 3) + C 1 so ∫ ln(2x + 3)dx = x ln(2x + 3) − x +
ln(2x + 3) + C
- u = ln(x^2 + 4), dv = dx, du = 2 x x^2 + 4 dx, v = x;
ln(x^2 + 4)dx = x ln(x^2 + 4) − 2
x^2 x^2 + 4 dx
but
x^2 x^2 + 4 dx =
x^2 + 4
dx = x − 2 tan−^1 x 2
- C 1 so ∫ ln(x^2 + 4)dx = x ln(x^2 + 4) − 2 x + 4 tan−^1 x 2
+ C
- u = sin−^1 x, dv = dx, du = 1/
1 − x^2 dx, v = x; ∫ sin−^1 x dx = x sin−^1 x −
x/
1 − x^2 dx = x sin−^1 x +
1 − x^2 + C
- u = cos−^1 (2x), dv = dx, du = − √^2 1 − 4 x^2
dx, v = x; ∫ cos−^1 (2x)dx = x cos−^1 (2x) +
∫ (^2) x √ 1 − 4 x^2
dx = x cos−^1 (2x) −
1 − 4 x^2 + C
- u = tan−^1 (2x), dv = dx, du = 2 1 + 4x^2 dx, v = x; ∫ tan−^1 (2x)dx = x tan−^1 (2x) −
2 x 1 + 4x^2 dx^ =^ x^ tan
− (^1) (2x) − 1 4 ln(1 + 4x
2 ) + C
- u = tan−^1 x, dv = x dx, du =
1 + x^2 dx, v =
x^2 ;
x tan−^1 x dx =
x^2 tan−^1 x −
x^2 1 + x^2 dx
but
∫ (^) x 2 1 + x^2 dx =
1 + x^2
dx = x − tan−^1 x + C 1 so ∫ x tan−^1 x dx =
x^2 tan−^1 x −
x +
tan−^1 x + C
- u = ex, dv = sin x dx, du = exdx, v = − cos x;
ex^ sin x dx = −ex^ cos x +
ex^ cos x dx.
For
ex^ cos x dx use u = ex, dv = cos x dx to get
ex^ cos x = ex^ sin x −
ex^ sin x dx so ∫ ex^ sin x dx = −ex^ cos x + ex^ sin x −
ex^ sin x dx,
ex^ sin x dx = ex(sin x − cos x) + C 1 ,
ex^ sin x dx =^1 2 ex(sin x − cos x) + C
- u = e^2 x, dv = cos 3x dx, du = 2e^2 xdx, v =
sin 3x; ∫ e^2 x^ cos 3x dx =^1 3 e^2 x^ sin 3x − 2 3
e^2 x^ sin 3x dx. Use u = e^2 x, dv = sin 3x dx to get
Exercise Set 8.2 321
∫ e^2 x^ sin 3x dx = − 1 3 e^2 x^ cos 3x +^2 3
e^2 x^ cos 3x dx so ∫ e^2 x^ cos 3x dx =
3 e
2 x (^) sin 3x +^2 9 e
2 x (^) cos 3x − 4 9
e^2 x^ cos 3x dx,
13 9
e^2 x^ cos 3x dx =
e^2 x(3 sin 3x + 2 cos 3x) + C 1 ,
e^2 x^ cos 3x dx =
e^2 x(3 sin 3x + 2 cos 3x) + C
- u = eax, dv = sin bx dx, du = aeaxdx, v = −
b cos bx (b = 0); ∫ eax^ sin bx dx = −
b e
ax (^) cos bx + a b
eax^ cos bx dx. Use u = eax, dv = cos bx dx to get ∫ eax^ cos bx dx =
b eax^ sin bx − a b
eax^ sin bx dx so ∫ eax^ sin bx dx = −
b eax^ cos bx + a b^2 eax^ sin bx − a^2 b^2
eax^ sin bx dx, ∫ eax^ sin bx dx = eax a^2 + b^2 (a^ sin^ bx^ −^ b^ cos^ bx) +^ C
- From Exercise 21 with a = − 3 , b = 5, x = θ, answer = e
− 3 θ √ 34
(−3 sin 5θ − 5 cos 5θ) + C
- u = sin(ln x), dv = dx, du = cos(ln x) x dx, v = x; ∫ sin(ln x)dx = x sin(ln x) −
cos(ln x)dx. Use u = cos(ln x), dv = dx to get ∫ cos(ln x)dx = x cos(ln x) +
sin(ln x)dx so ∫ sin(ln x)dx = x sin(ln x) − x cos(ln x) −
sin(ln x)dx, ∫ sin(ln x)dx =^1 2 x[sin(ln x) − cos(ln x)] + C
- u = cos(ln x), dv = dx, du = −
x sin(ln x)dx, v = x; ∫ cos(ln x)dx = x cos(ln x) +
sin(ln x)dx. Use u = sin(ln x), dv = dx to get ∫ sin(ln x)dx = x sin(ln x) −
cos(ln x)dx so ∫ cos(ln x)dx = x cos(ln x) + x sin(ln x) −
cos(ln x)dx, ∫ cos(ln x)dx =
x[cos(ln x) + sin(ln x)] + C
Exercise Set 8.2 323
- u = ln(x + 3), dv = dx, du =
x + 3 dx, v = x; ∫ (^2)
− 2
ln(x + 3)dx = x ln(x + 3)
] 2
− 2
− 2
x x + 3 dx = 2 ln 5 + 2 ln 1 −
− 2
[
x + 3
]
dx
= 2 ln 5 − [x − 3 ln(x + 3)]
] 2
− 2
= 2 ln 5 − (2 − 3 ln 5) + (− 2 − 3 ln 1) = 5 ln 5 − 4
- u = sin−^1 x, dv = dx, du =
1 − x^2
dx, v = x;
∫ (^1) / 2
0
sin−^1 x dx = x sin−^1 x
] 1 / 2
0
0
√^ x 1 − x^2
dx =^1 2 sin−^1 2
1 − x^2
] 1 / 2
0
=^1 2
( (^) π 6
− 1 = π 12
- u = sec−^1
θ, dv = dθ, du = 1 2 θ
θ − 1
dθ, v = θ;
∫ (^4)
2
sec−^1
θdθ = θ sec−^1
θ
] 4
2
2
θ − 1
dθ = 4 sec−^1 2 − 2 sec−^1
θ − 1
] 4
2 = 4
( (^) π 3
( (^) π 4
5 π 6
- u = sec−^1 x, dv = x dx, du =
x
x^2 − 1
dx, v =
x^2 ;
∫ (^2)
1
x sec−^1 x dx =^1 2 x^2 sec−^1 x
] 2
1
1
√^ x x^2 − 1
dx
=^1
[(4)(π/3) − (1)(0)] − 1 2
x^2 − 1
] 2
1
= 2π/ 3 −
- u = x, dv = sin 4x dx, du = dx, v = − 1 4 cos 4x; ∫ (^) π/ 2
0
x sin 4x dx = − 1 4 x cos 4x
]π/ 2
0
+^1
∫ (^) π/ 2
0
cos 4x dx = −π/8 +^1 16 sin 4x
]π/ 2
0
= −π/ 8
∫ (^) π
0
(x + x cos x)dx =^1 2 x^2
]π
0
∫ (^) π
0
x cos x dx = π
2 2
∫ (^) π
0
x cos x dx;
u = x, dv = cos x dx, du = dx, v = sin x ∫ (^) π
0
x cos x dx = x sin x
]π
0
∫ (^) π
0
sin x dx = cos x
]π
0
= −2 so
∫ (^) π
0
(x + x cos x)dx = π^2 / 2 − 2
324 Chapter 8
- u = tan−^1
x, dv =
xdx, du =
x( 1 +x) dx, v =
x^3 /^2 ; ∫ (^3)
1
x tan−^1
xdx =
x^3 /^2 tan−^1
x
] 3
1
1
x 1 + x dx
x^3 /^2 tan−^1
x
] 3
1
1
[
1 + x
]
dx
[
x^3 /^2 tan−^1
x −
x +
ln |1 + x|
] 3
1
3 π − π/ 2 − 2 + ln 2)/ 3
- u = ln(x^2 + 1), dv = dx, du = 2 x x^2 + 1 dx, v = x; ∫ (^2)
0
ln(x^2 + 1)dx = x ln(x^2 + 1)
] 2
0
0
2 x^2 x^2 + 1 dx = 2 ln 5 − 2
0
x^2 + 1
dx
= 2 ln 5 − 2(x − tan−^1 x)
] 2
0
= 2 ln 5 − 4 + 2 tan−^1
- t =
x, t^2 = x, dx = 2t dt
(a)
e
√x dx = 2
tet^ dt; u = t, dv = etdt, du = dt, v = et, ∫ e
√x dx = 2tet^ − 2
et^ dt = 2(t − 1)et^ + C = 2(
x − 1)e
√x
(b)
cos
x dx = 2
t cos t dt; u = t, dv = cos tdt, du = dt, v = sin t, ∫ cos
x dx = 2t sin t − 2
sin tdt = 2t sin t + 2 cos t + C = 2
x sin
x + 2 cos
x + C
- Let q 1 (x), q 2 (x), q 3 (x) denote successive antiderivatives of q(x), so that q′ 3 (x) = q 2 (x), q 2 ′(x) = q 1 (x), q′ 1 (x) = q(x). Let p(x) = ax^2 + bx + c.
Repeated Repeated Differentiation Antidifferentiation ax^2 + bx + c q(x)
2 ax + b q 1 (x) − 2 a q 2 (x)
0 q 3 (x)
Then
p(x)q(x) dx = (ax^2 + bx + c)q 1 (x) − (2ax + b)q 2 (x) + 2aq 3 (x) + C. Check:
d dx [(ax
(^2) +bx + c)q 1 (x) − (2ax + b)q 2 (x) + 2aq 3 (x)]
= ( 2ax + b)q 1 (x) + (ax^2 + bx + c)q(x) − 2 aq 2 (x) − (2ax + b)q 1 (x) + 2aq 2 (x) = p(x)q(x)
326 Chapter 8
- Repeated Repeated Differentiation Antidifferentiation x^3
2 x + 1
3 x^2
(2x + 1)^3 /^2 − 6 x
(2x + 1)^5 /^2
6
(2x + 1)^7 /^2 − 0
(2x + 1)^9 /^2
x^3
2 x + 1 dx =^1 3 x^3 (2x + 1)^3 /^2 − 1 5 x^2 (2x + 1)^5 /^2 +^2 35 x(2x + 1)^7 /^2 − 2 315 (2x + 1)^9 /^2 + C
- (a) A =
∫ (^) e
1
ln x dx = (x ln x − x)
]e
1
(b) V = π
∫ (^) e
1
(ln x)^2 dx = π
[
(x(ln x)^2 − 2 x ln x + 2x)
]e 1 = π(e − 2)
48. A =
∫ (^) π/ 2
0
(x − x sin x)dx =
x^2
]π/ 2
0
∫ (^) π/ 2
0
x sin x dx = π^2 8 − (−x cos x + sin x)
]π/ 2
0
= π^2 / 8 − 1
- V = 2π
∫ (^) π
0
x sin x dx = 2π(−x cos x + sin x)
]π
0
= 2π^2
- V = 2π
∫ (^) π/ 2
0
x cos x dx = 2π(cos x + x sin x)
]π/ 2
0
= π(π − 2)
- distance =
0
t^2 e−tdt; u = t^2 , dv = e−tdt, du = 2tdt, v = −e−t,
distance = −t^2 e−t
] 5
0
0
te−tdt; u = 2t, dv = e−tdt, du = 2dt, v = −e−t,
distance = − 25 e−^5 − 2 te−t
] 5
0
0
e−tdt = − 25 e−^5 − 10 e−^5 − 2 e−t
] 5
0
= − 25 e−^5 − 10 e−^5 − 2 e−^5 + 2 = − 37 e−^5 + 2
Exercise Set 8.2 327
- u = 2t, dv = sin(kωt)dt, du = 2dt, v = −
kω cos(kωt); the integrand is an even function of t so ∫ (^) π/ω
−π/ω
t sin(kωt) dt = 2
∫ (^) π/ω
0
t sin(kωt) dt = −
kω t cos(kωt)
]π/ω
0
∫ (^) π/ω
0
kω cos(kωt) dt
2 π(−1)k+ kω^2
k^2 ω^2 sin(kωt)
]π/ω
0
2 π(−1)k+ kω^2
- (a)
sin^3 x dx = − 1 3 sin^2 x cos x +^2 3
sin x dx = − 1 3 sin^2 x cos x − 2 3 cos x + C
(b)
sin^4 x dx = − 1 4 sin^3 x cos x +^3 4
sin^2 x dx,
sin^2 x dx = − 1 2 sin x cos x +^1 2 x + C 1 so ∫ (^) π/ 4
0
sin^4 x dx =
[
sin^3 x cos x −
sin x cos x +
x
]π/ 4
0
= −
2)^3 (1/
- (a)
cos^5 x dx =
cos^4 x sin x +
cos^3 x dx =
cos^4 x sin x +
[
cos^2 x sin x +
sin x
]
+ C
cos^4 x sin x +
cos^2 x sin x +
sin x + C
(b)
cos^6 x dx =
cos^5 x sin x +
cos^4 x dx
cos^5 x sin x +
[
cos^3 x sin x +
cos^2 x dx
]
cos^5 x sin x +
cos^3 x sin x +
[
cos x sin x +
x
]
+ C,
[
cos^5 x sin x +
cos^3 x sin x +
cos x sin x +
x
]π/ 2
0
= 5π/ 32
- u = sinn−^1 x, dv = sin x dx, du = (n − 1) sinn−^2 x cos x dx, v = − cos x; ∫ sinn^ x dx = − sinn−^1 x cos x + (n − 1)
sinn−^2 x cos^2 x dx
= − sinn−^1 x cos x + (n − 1)
sinn−^2 x (1 − sin^2 x)dx
= − sinn−^1 x cos x + (n − 1)
sinn−^2 x dx − (n − 1)
sinn^ x dx,
n
sinn^ x dx = − sinn−^1 x cos x + (n − 1)
sinn−^2 x dx, ∫ sinn^ x dx = −
n sinn−^1 x cos x + n − 1 n
sinn−^2 x dx
Exercise Set 8.2 329
- u = x, dv = f ′′(x)dx, du = dx, v = f ′(x);
∫ (^1)
− 1
x f ′′(x)dx = xf ′(x)
] 1
− 1
− 1
f ′(x)dx
= f ′(1) + f ′(−1) − f (x)
] 1
− 1
= f ′(1) + f ′(−1) − f (1) + f (−1)
- (a) u = f (x), dv = dx, du = f ′(x), v = x; ∫ (^) b
a
f (x) dx = xf (x)
]b
a
∫ (^) b
a
xf ′(x) dx = bf (b) − af (a) −
∫ (^) b
a
xf ′(x) dx
(b) Substitute y = f (x), dy = f ′(x) dx, x = a when y = f (a), x = b when y = f (b), ∫ (^) b
a
xf ′(x) dx =
∫ (^) f (b)
f (a)
x dy =
∫ (^) f (b)
f (a)
f −^1 (y) dy
(c) From a = f −^1 (α) and b = f −^1 (β) we get bf (b) − af (a) = βf −^1 (β) − αf −^1 (α); then ∫ (^) β
α
f −^1 (x) dx =
∫ (^) β
α
f −^1 (y) dy =
∫ (^) f (b)
f (a)
f −^1 (y) dy,
which, by Part (b), yields
a
b
a = f –1( a ) b = f –1( b )
x
y
A 1 A 2
∫ (^) β
α
f −^1 (x) dx = bf (b) − af (a) −
∫ (^) b
a
f (x) dx
= βf −^1 (β) − αf −^1 (α) −
∫ (^) f −^1 (β)
f −^1 (α)
f (x) dx
Note from the figure that A 1 =
∫ (^) β
α
f −^1 (x) dx, A 2 =
∫ (^) f −^1 (β)
f −^1 (α)
f (x) dx, and
A 1 + A 2 = βf −^1 (β) − αf −^1 (α), a ”picture proof”.
- (a) Use Exercise 60(c); ∫ (^1) / 2
0
sin−^1 x dx =
sin−^1
− 0 ·sin−^1 0 −
∫ (^) sin−^1 (1/2)
sin−^1 (0)
sin x dx =
sin−^1
∫ (^) π/ 6
0
sin x dx
(b) Use Exercise 60(b); ∫ (^) e^2
e
ln x dx = e^2 ln e^2 − e ln e −
∫ (^) ln e^2
ln e
f −^1 (y) dy = 2e^2 − e −
1
ey^ dy = 2e^2 − e −
1
ex^ dx
- (a)
u dv = uv −
v du = x(sin x + C 1 ) + cos x − C 1 x + C 2 = x sin x + cos x + C 2 ;
the constant C 1 cancels out and hence plays no role in the answer.
(b) u(v + C 1 ) −
(v + C 1 )du = uv + C 1 u −
v du − C 1 u = uv −
v du
330 Chapter 8
- u = ln(x + 1), dv = dx, du = dx ∫ x^ + 1^ , v^ =^ x^ + 1; ln(x + 1) dx =
u dv = uv −
v du = (x + 1) ln(x + 1) −
dx = (x + 1) ln(x + 1) − x + C
- u = ln(2x + 3), dv = dx, du = 2 dx 2 x + 3 , v = x +
ln(2x + 3) dx =
u dv = uv −
v du = (x +
) ln(2x + 3) −
dx
2 (2x^ + 3) ln(2x^ + 3)^ −^ x^ +^ C
- u = tan−^1 x, dv = x dx, du = 1 1 + x^2 dx, v =^1 2 (x^2 + 1) ∫ x tan−^1 x dx =
u dv = uv −
v du =
2 (x
(^2) + 1) tan− (^1) x − 1 2
dx
=^1 2 (x^2 + 1) tan−^1 x − 1 2 x + C
- u =
ln x , dv =
x dx, du = −
x(ln x)^2 dx, v = ln x ∫ 1 x ln x dx = 1 +
x ln x dx.
This seems to imply that 1 = 0, but recall that both sides represent a function plus an arbitrary constant; these two arbitrary constants will take care of the 1.
EXERCISE SET 8.
- u = cos x, −
u^5 du = −
6 cos
(^6) x + C 2. u = sin 3x,^1 3
u^4 du =
15 sin
(^5 3) x + C
- u = sin ax,^1 a
u du = 1 2 a sin^2 ax + C, a = 0
cos^2 3 x dx =^1 2
(1 + cos 6x)dx =^1 2 x +^1 12 sin 6x + C
sin^2 5 θ dθ =^1 2
(1 − cos 10θ)dθ =^1 2 θ − 1 20 sin 10θ + C
cos^3 at dt =
(1 − sin^2 at) cos at dt
cos at dt −
sin^2 at cos at dt =^1 a sin at − 1 3 a sin^3 at + C (a = 0)
cos^5 θdθ =
(1 − sin^2 θ)^2 cos θdθ =
(1 − 2 sin^2 θ + sin^4 θ) cos θdθ
= sin θ −
sin^3 θ +
sin^5 θ + C
332 Chapter 8
∫ (^) π
−π
cos^2 5 θ dθ =
∫ (^) π
−π
(1 + cos 10θ)dθ =
θ +
sin 10θ
)]π
−π
= π
∫ (^) π/ 6
0
sin 2x cos 4x dx =
∫ (^) π/ 6
0
(sin 6x − sin 2x)dx =
[
cos 6x +
cos 2x
]π/ 6
0 = [(− 1 /12)(−1) + ( 1/4)(1/2)] − [− 1 /12 + 1/4] = 1/ 24
∫ (^2) π
0
sin^2 kx dx =
∫ (^2) π
0
(1 − cos 2kx)dx =
x −
2 k sin 2kx
)] 2 π
0
= π −
4 k sin 4πk^ (k^ = 0)
tan(3x + 1) + C 24. −
ln | cos 5x| + C
- u = e−^2 x, du = − 2 e−^2 x^ dx; −
tan u du =
2 ln^ |^ cos^ u|^ +^ C^ =
2 ln^ |^ cos(e
− 2 x)| + C
3 ln^ |^ sin 3x|^ +^ C^ 27.^
2 ln^ |^ sec 2x^ + tan 2x|^ +^ C
- u =
x, du =
x dx;
2 sec u du = 2 ln | sec u + tan u| + C = 2 ln
sec
x + tan
x
+ C
- u = tan x,
u^2 du =^1 3 tan^3 x + C
tan^5 x(1 + tan^2 x) sec^2 x dx =
(tan^5 x + tan^7 x) sec^2 x dx =
tan^6 x +
tan^8 x + C
tan^3 4 x(1 + tan^2 4 x) sec^2 4 x dx =
(tan^3 4 x + tan^5 4 x) sec^2 4 x dx = 1 16 tan^4 4 x +^1 24 tan^6 4 x + C
tan^4 θ(1 + tan^2 θ) sec^2 θ dθ =
tan^5 θ +
tan^7 θ + C
sec^4 x(sec^2 x − 1) sec x tan x dx =
(sec^6 x − sec^4 x) sec x tan x dx =^1 7 sec^7 x − 1 5 sec^5 x + C
(sec^2 θ − 1)^2 sec θ tan θdθ =
(sec^4 θ − 2 sec^2 θ + 1) sec θ tan θdθ =
sec^5 θ −
sec^3 θ + sec θ + C
(sec^2 x − 1)^2 sec x dx =
(sec^5 x − 2 sec^3 x + sec x)dx =
sec^5 x dx − 2
sec^3 x dx +
sec x dx
=^1
sec^3 x tan x +^3 4
sec^3 x dx − 2
sec^3 x dx + ln | sec x + tan x|
=^1
sec^3 x tan x − 5 4
[
sec x tan x +^1 2 ln | sec x + tan x|
]
sec^3 x tan x −
sec x tan x +
ln | sec x + tan x| + C
Exercise Set 8.3 333
[sec^2 (x/2) − 1] sec^3 (x/2)dx =
[sec^5 (x/2) − sec^3 (x/2)]dx
= 2
[∫
sec^5 u du −
sec^3 u du
]
(u = x/2)
[(
sec^3 u tan u +^3 4
sec^3 u du
sec^3 u du
]
(equation (20))
2 sec
(^3) u tan u − 1 2
sec^3 u du
sec^3 u tan u −
sec u tan u −
ln | sec u + tan u| + C (equation (20), (22))
sec^3 x 2 tan x 2
sec x 2 tan x 2
ln
∣sec x 2
∣ + C
sec^2 2 t(sec 2t tan 2t)dt =
sec^3 2 t + C 38.
sec^4 x(sec x tan x)dx =
sec^5 x + C
sec^4 x dx =
(1 + tan^2 x) sec^2 x dx =
(sec^2 x + tan^2 x sec^2 x)dx = tan x +^1 3 tan^3 x + C
- Using equation (20), ∫ sec^5 x dx =
sec^3 x tan x +
sec^3 x dx
sec^3 x tan x +
sec x tan x +
ln | sec x + tan x| + C
- Use equation (19) to get
tan^4 x dx =
3 tan
(^3) x − tan x + x + C
- u = 4x, use equation (19) to get
1 4
tan^3 u du =
[
tan^2 u + ln | cos u|
]
+ C =
tan^2 4 x +
ln | cos 4x| + C
tan x(1 + tan^2 x) sec^2 x dx =
3 tan
3 / (^2) x +^2 7 tan
7 / (^2) x + C
sec^1 /^2 x(sec x tan x)dx =^2 3 sec^3 /^2 x + C
∫ (^) π/ 6
0
(sec^2 2 x − 1)dx =
[
2 tan 2x^ −^ x
]π/ 6
0
3 / 2 − π/ 6
∫ (^) π/ 6
0
sec^2 θ(sec θ tan θ)dθ =^1 3 sec^3 θ
]π/ 6
0
3)^3 − 1 /3 = 8
- u = x/2,
∫ (^) π/ 4
0
tan^5 u du =
[
tan^4 u − tan^2 u − 2 ln | cos u|
]π/ 4
0
= 1/ 2 − 1 − 2 ln(1/
- = − 1 /2 + ln 2
Exercise Set 8.3 335
- V = π
∫ (^) π/ 4
0
(cos^2 x − sin^2 x)dx = π
∫ (^) π/ 4
0
cos 2x dx =
π sin 2x
]π/ 4
0
= π/ 2
- V = π
∫ (^) π
0
sin^2 x dx = π 2
∫ (^) π
0
(1 − cos 2x)dx = π 2
x −
2 sin 2x
)]π
0
= π^2 / 2
- With 0 < α < β, D = Dβ −Dα =
L
2 π
∫ (^) β
α
sec x dx =
L
2 π ln | sec x + tan x|
]β
α
L
2 π ln
∣∣^ sec^ β^ + tan^ β sec α + tan α
- (a) D =
2 π ln(sec 25
◦ (^) + tan 25◦) = 7.18 cm
(b) D =
2 π ln
∣∣^ sec 50
◦ (^) + tan 50◦ sec 30◦^ + tan 30◦
∣∣ = 7.34 cm
- (a)
csc x dx =
sec(π/ 2 − x)dx = − ln | sec(π/ 2 − x) + tan(π/ 2 − x)| + C = − ln | csc x + cot x| + C
(b) − ln | csc x + cot x| = ln
| csc x + cot x| = ln | csc x − cot x| | csc^2 x − cot^2 x| = ln | csc x − cot x|,
− ln | csc x + cot x| = − ln
∣∣^1
sin x
cos x sin x
∣∣ = ln
∣∣^ sin^ x 1 + cos x
= ln
∣∣^ 2 sin(x/2) cos(x/2) 2 cos^2 (x/2)
∣∣ = ln | tan(x/2)|
- sin x + cos x =
[
- sin x + ( 1/
- cos x
]
2 [sin x cos(π/4) + cos x sin(π/4)] =
2 sin(x + π/4), ∫ dx sin x + cos x
= √^1
csc(x + π/4)dx = − √^1 2
ln | csc(x + π/4) + cot(x + π/4)| + C
= − √^1
ln
2 + cos x − sin x sin x + cos x
+ C
- a sin x + b cos x =
a^2 + b^2
[
a √ a^2 + b^2
sin x + b √ a^2 + b^2
cos x
]
a^2 + b^2 (sin x cos θ + cos x sin θ)
where cos θ = a/
a^2 + b^2 and sin θ = b/
a^2 + b^2 so a sin x + b cos x =
a^2 + b^2 sin(x + θ)
and
dx a sin x + b cos x
= √^1
a^2 + b^2
csc(x + θ)dx = − √^1 a^2 + b^2
ln | csc(x + θ) + cot(x + θ)| + C
= − √^1
a^2 + b^2
ln
a^2 + b^2 + a cos x − b sin x a sin x + b cos x
+ C
336 Chapter 8
- (a)
∫ (^) π/ 2
0
sinn^ x dx = −
n sinn−^1 x cos x
]π/ 2
0
n − 1 n
∫ (^) π/ 2
0
sinn−^2 x dx = n − 1 n
∫ (^) π/ 2
0
sinn−^2 x dx
(b) By repeated application of the formula in Part (a) ∫ (^) π/ 2
0
sinn^ x dx =
n − 1 n
n − 3 n − 2
) ∫ (^) π/ 2
0
sinn−^4 x dx
n − 1 n
n − 3 n − 2
n − 5 n − 4
) ∫ (^) π/ 2
0
dx, n even ( n − 1 n
n − 3 n − 2
n − 5 n − 4
) ∫ (^) π/ 2
0
sin x dx, n odd
1 · 3 · 5 · · · (n − 1) 2 · 4 · 6 · · · n · π 2 , n even
2 · 4 · 6 · · · (n − 1) 3 · 5 · 7 · · · n , n odd
- (a)
∫ (^) π/ 2
0
sin^3 x dx =
3 (b)
∫ (^) π/ 2
0
sin^4 x dx =
2 · 4 ·^
π 2 = 3π/^16
(c)
∫ (^) π/ 2
0
sin^5 x dx =^2 ·^4 3 · 5 = 8/ 15 (d)
∫ (^) π/ 2
0
sin^6 x dx =^1 ·^3 ·^5 2 · 4 · 6 · π 2 = 5π/ 32
- Similar to proof in Exercise 64.
EXERCISE SET 8.
- x = 2 sin θ, dx = 2 cos θ dθ,
4
cos^2 θ dθ = 2
(1 + cos 2θ)dθ = 2θ + sin 2θ + C
= 2θ + 2 sin θ cos θ + C = 2 sin−^1 (x/2) +
x
4 − x^2 + C
- x =
sin θ, dx =
cos θ dθ,
1 2
cos^2 θ dθ =^1 4
(1 + cos 2θ)dθ =^1 4 θ +^1 8 sin 2θ + C
θ +
sin θ cos θ + C =
sin−^1 2 x +
x
1 − 4 x^2 + C
- x = 3 sin θ, dx = 3 cos θ dθ,
9
sin^2 θ dθ =
(1 − cos 2θ)dθ =
θ −
sin 2θ + C =
θ −
sin θ cos θ + C
=^9 2 sin−^1 (x/3) − 1 2 x
9 − x^2 + C
- x = 4 sin θ, dx = 4 cos θ dθ, 1 16
sin^2 θ
dθ = 1 16
csc^2 θ dθ = − 1 16 cot θ + C = −
16 − x^2 16 x
+ C
