Integration by parts 421
Problem 2. Find
∫
3 te^2 tdt.
Letu= 3 t, from which,
du
dt
=3, i.e. du=3dtand
let dv=e^2 tdt, from which,v=
∫
e^2 tdt=
1
2
e^2 t
Substituting into
∫
udv=uv−
∫
vdugives:
∫
3 te^2 tdt=( 3 t)
(
1
2
e^2 t
)
−
∫ (
1
2
e^2 t
)
(3dt)
=
3
2
te^2 t−
3
2
∫
e^2 tdt
=
3
2
te^2 t−
3
2
(
e^2 t
2
)
+c
Hence
∫
3 te^2 tdt=^32 e^2 t
(
t−^12
)
+c,
which may be checked by differentiating.
Problem 3. Evaluate
∫ π
2
0
2 θsinθdθ.
Letu= 2 θ, from which,
du
dθ
=2, i.e. du=2dθand let
dv=sinθdθ, from which,
v=
∫
sinθdθ=−cosθ
Substituting into
∫
udv=uv−
∫
vdugives:
∫
2 θsinθdθ=( 2 θ)(−cosθ)−
∫
(−cosθ)(2dθ)
=− 2 θcosθ+ 2
∫
cosθdθ
=− 2 θcosθ+2sinθ+c
Hence
∫ π
2
0
2 θsinθdθ
=[− 2 θcosθ+2sinθ]
π
2
0
=
[
− 2
(π
2
)
cos
π
2
+2sin
π
2
]
−[0+2sin0]
=(− 0 + 2 )−( 0 + 0 )= 2
sincecos
π
2
=0andsin
π
2
= 1
Problem 4. Evaluate
∫ 1
0
5 xe^4 xdx, correct to
3 significant figures.
Letu= 5 x, from which
du
dx
=5, i.e. du=5dxand
let dv=e^4 xdx, from which,v=
∫
e^4 xdx=^14 e^4 x.
Substituting into
∫
udv=uv−
∫
vdugives:
∫
5 xe^4 xdx=( 5 x)
(
e^4 x
4
)
−
∫ (
e^4 x
4
)
(5dx)
=
5
4
xe^4 x−
5
4
∫
e^4 xdx
=
5
4
xe^4 x−
5
4
(
e^4 x
4
)
+c
=
5
4
e^4 x
(
x−
1
4
)
+c
Hence
∫ 1
0
5 xe^4 xdx
=
[
5
4
e^4 x
(
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 sinxdx.
Letu=x^2 , from which,
du
dx
= 2 x,i.e.du= 2 xdx,and
let dv=sinxdx, from which,
v=
∫
sinxdx=−cosx
Substituting into
∫
udv=uv−
∫
vdugives:
∫
x^2 sinxdx=(x^2 )(−cosx)−
∫
(−cosx)( 2 xdx)
=−x^2 cosx+ 2
[∫
xcosxdx
]