Integration by parts 421
Problem 2. Find
3t e
2t dt .
Let u =3t , from which,
du
dt
= 3, i.e. du = 3 dt and
let dv = e 2t dt , from which, v =
e 2t dt =
1
2
e 2t
Substituting into
u dv = uv −
v du gives:
3t e
2t dt = (3t )
1
2
e
2t
−
1
2
e
2t
(3 dt )
=
3
2
t e
2t
−
3
2
e
2t dt
=
3
2
t e
2t
−
3
2
e 2t
2
+ c
Hence
3t e
2t dt =
3
2 e
2t
t −
1
2
+ c,
which may be checked by differentiating.
Problem 3. Evaluate
π
2
0
2θ sin θ dθ.
Let u = 2θ, from which,
du
dθ
= 2, i.e. du =2 dθ and let
dv = sin θ dθ, from which,
v =
sin θ dθ =−cos θ
Substituting into
u dv = uv −
v du gives:
2θ sin θ dθ = (2θ)(−cos θ) −
(−cos θ)(2 dθ)
= −2θ cos θ + 2
cos θ dθ
= −2θ cos θ + 2 sin θ + c
Hence
π
2
0
2θ sin θ dθ
= [−2θ cos θ + 2 sinθ]
π
2
0
=
−2
π
2
cos
π
2
+ 2 sin
π
2
− [0 + 2 sin0]
= (−0 + 2) − (0 + 0) = 2
sincecos
π
2
= 0 and sin
π
2
= 1
Problem 4. Evaluate
1
0
5xe
4x dx, correct to
3 significant figures.
Let u =5x, from which
du
dx
= 5, i.e. du = 5 dx and
let dv = e 4x dx, from which, v =
e 4x dx =
1
4 e 4x .
Substituting into
u dv = uv −
v du gives:
5xe
4x dx = (5x)
e 4x
4
−
e 4x
4
(5 dx)
=
5
4
xe
4x
−
5
4
e
4x dx
=
5
4
xe
4x
−
5
4
e 4x
4
+ c
=
5
4
e
4x
x −
1
4
+ c
Hence
1
0
5xe
4x dx
=
5
4
e
4x
x −
1
4
1
0
=
5
4
e
4
1 −
1
4
−
5
4
e
0
0 −
1
4
=
15
16
e
4
−
−
5
16
= 51.186 + 0.313 = 51.499 = 51.5,
correct to 3 significant figures
Problem 5. Determine
x 2 sin x dx.
Let u = x 2 , from which,
du
dx
= 2x, i.e. du =2x dx, and
let dv = sin x dx, from which,
v =
sin x dx = −cos x
Substituting into
u dv = uv −
v du gives:
x
2 sin x dx = (x
2
)(−cos x) −
(−cos x)(2x dx)
= −x
2 cos x + 2
x cos x dx