Higher Engineering Mathematics

INTEGRATION BY PARTS 419

H

=

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=(2θ)(−cosθ)−

∫

(−cosθ)(2 dθ)

=− 2 θcosθ+ 2

∫

cosθdθ

=− 2 θcosθ+2 sinθ+c

Hence

∫π

2

0

2 θsinθdθ

=[− 2 θcosθ+2 sinθ]

π

2

0

=

[

− 2

(π

2

)

cos

π

2

+2 sin

π

2

]

−[0+2 sin 0]

=(− 0 +2)−(0+0)= 2

since cos

π

2

=0 and sin

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=(5x)

(

e^4 x

4

)

−

∫ (

e^4 x

4

)

(5 dx)

=

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)(2xdx)

=−x^2 cosx+ 2

[∫

xcosxdx

]

The integral,

∫

xcosxdx, is not a ‘standard integral’

and it can only be determined by using the integration

by parts formula again.