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Millersville University
Department of Mathematics
MATH 211, Calculus II
July 15, 2008
Please evaluate the following definite and indefinite integrals.
∫ (^) e
1
x 3 ln x dx
Using integration by parts we will let
u = ln x v =
x 4
du =
x
dx dv = x 3 dx
then ∫ (^) e
1
x 3 ln x dx =
x 4 ln x
∣ ∣ ∣ ∣
e
1
∫ (^) e
1
x 3 dx
x 4 ln x
∣ ∣ ∣ ∣
e
1
x 4
∣ ∣ ∣ ∣
e
1
=
( 1
4
e 4 ln e −
ln 1
) −
( 1
16
e 4 −
)
e 4
∫ sin − 1 x dx
Using integration by parts we will let
u = sin−^1 x v = x
du =
1 − x^2
dx dv = dx
then (^) ∫
sin − 1 x dx = x sin − 1 x −
∫ 1 √ 1 − x^2
dx.
Now we make the substitution w = 1 − x 2 and − 1 2 dw^ =^ x dx^ to obtain ∫ sin − 1 x dx = x sin − 1 x −
∫ ( −
) w − 1 / 2 dw
= x sin−^1 x +
∫ w−^1 /^2 dw
= x sin − 1 x + w 1 / 2
= x sin − 1 x +
1 − x^2 + C
∫ 4 x sec 2 2 x dx
Using integration by parts we will let
u = 4x v =
tan 2x
du = 4 dx dv = sec 2 2 x dx
then ∫ 4 x sec 2 2 x dx = 2 x tan 2x −
∫ 2 tan 2x dx
= 2 x tan 2x −
∫ 2 sin 2x
cos 2x
dx
Now we make the substitution w = cos 2x and −dw = 2 sin 2x dx to obtain ∫ 4 x sec 2 2 x dx = 2 x tan 2x +
∫ 1
w
dw
= 2 x tan 2x + ln |w| + C
= 2 x tan 2x + ln | cos 2x| + C
∫ p 4 e −p dp
Using the method of tabular integration by parts we obtain
u dv sign e −p
p^4 −e−p^ + 4 p^3 e−p^ − 12 p^2 −e−p^ + 24 p e −p − 24 −e −p
0 e −p −
and therefore ∫ p 4 e −p dp = −p 4 e −p − 4 p 3 e −p − 12 p 2 e −p − 24 pe −p − 24 e −p
∫ (^1) /√ 2
0
2 x sin − 1 (x 2 ) dx
Using integration by substitution we will let u = x^2 and du = 2x dx. Then
∫ (^1) /√ 2
0
2 x sin − 1 (x 2 ) dx =
∫ (^) u(1/√2)
u(0)
sin − 1 u du
∫ (^1)
0
u 1 / 2 − u 3 / 2 du
u 3 / 2 −
u 5 / 2
∣ ∣ ∣ ∣
1
0
=
∫ (^) π/ 3
0
x tan 2 x dx
We will start this problem by making use of a trigonometric identity, 1+tan 2 x = sec 2 x.
∫ (^) π/ 3
0
x tan 2 x dx =
∫ (^) π/ 3
0
x(−1 + sec 2 x) dx
∫ (^) π/ 3
0
x dx +
∫ (^) π/ 3
0
x sec 2 x dx
x 2
∣ ∣ ∣ ∣
π/ 3
0
∫ (^) π/ 3
0
x sec 2 x dx
π 2
∫ (^) π/ 3
0
x sec 2 x dx
Now we will use integration by parts on the remaining integral.
u = x v = tan x du = dx dv = sec 2 x dx
∫ (^) π/ 3
0
x tan 2 x dx = −
π^2
18
π/ 3 0 −
∫ (^) π/ 3
0
tan x dx
π^2
18
π
3
3 − ln(cos x)|
π/ 3 0
π^2
18
π
− ln(1/2)
π^2
18
π
≈ 0. 572341
∫ sin(ln x) dx
Using integration by parts we will make the following assignments.
u = sin(ln x) v = x
du = cos(ln x) ·
x
dx dv = dx
Applying the integration by parts formula produces:
∫ sin(ln x) dx = x sin(ln x) −
∫ cos(ln x) dx
We must use integration by parts again on the remaining integral.
u = cos(ln x) v = x
du = − sin(ln x) ·
x
dx dv = dx
Once again the integration by parts formula
produces:
∫ sin(ln x) dx = x sin(ln x) −
( x cos(ln x) +
∫ sin(ln x) dx
)
= x sin(ln x) − x cos(ln x) −
∫ sin(ln x) dx
Notice that the remaining integral is the same as the original integral, thus we will add
this expression to both sides of the equation.
∫ sin(ln x) dx = x sin(ln x) − x cos(ln x) ∫ sin(ln x) dx =
(x sin(ln x) − x cos(ln x)) + C
