428 Higher Engineering Mathematics

From equation (2),

∫

t^3 costdt=I 3 =t^3 sint+ 3 t^2 cost− 3 ( 2 )I 1

and

I 1 =t^1 sint+ 1 t^0 cost− 1 ( 0 )In− 2

=tsint+cost

Hence

∫

t^3 costdt=t^3 sint+ 3 t^2 cost

− 3 ( 2 )[tsint+cost]

=t^3 sint+ 3 t^2 cost− 6 tsint−6cost

Thus

∫ 2

1

4 t^3 costdt

=[4(t^3 sint+ 3 t^2 cost− 6 tsint−6cost)]^21

=[4(8sin2+12cos2−12sin2−6cos2)]

−[4(sin1+3cos1−6sin1−6cos1)]

=(− 24. 53628 )−(− 23. 31305 )

=− 1. 223

Problem 5. Determine a reduction formula

for

∫π

0 x

ncosxdxand hence evaluate

∫π

0 x

(^4) cosxdx, correct to 2 decimal places.
From equation (2),
In=xnsinx+nxn−^1 cosx−n(n− 1 )In− 2.
hence
∫π
0
xncosxdx=[xnsinx+nxn−^1 cosx]π 0
−n(n− 1 )In− 2
=[(πnsinπ+nπn−^1 cosπ)
−( 0 + 0 )]−n(n− 1 )In− 2
=−nπn−^1 −n(n−1)In− 2
Hence
∫π
0
x^4 cosxdx=I 4
=− 4 π^3 − 4 ( 3 )I 2 sincen= 4
Whenn=2,
∫π
0
x^2 cosxdx=I 2 =− 2 π^1 − 2 ( 1 )I 0
and I 0 =
∫π
0
x^0 cosxdx

∫π
0
cosxdx
=[sinx]π 0 = 0
Hence
∫π
0
x^4 cosxdx=− 4 π^3 − 4 ( 3 )[− 2 π− 2 ( 1 )( 0 )]
=− 4 π^3 + 24 πor−48.63,
correct to 2 decimal places.
(b)
∫
xnsinxdx
LetIn=
∫
xnsinxdx
Using integration by parts, ifu=xnthen
du
dx
=nxn−^1 and if dv=sinxdxthen
v=
∫
sinxdx=−cosx. Hence
∫
xnsinxdx
=In=xn(−cosx)−
∫
(−cosx)nxn−^1 dx
=−xncosx+n
∫
xn−^1 cosxdx
Using integration by parts again, withu=xn−^1 , from
which,
du
dx
=(n− 1 )xn−^2 and dv=cosx, from which,
v=
∫
cosxdx=sinx. Hence
In=−xncosx+n
[
xn−^1 (sinx)
−
∫
(sinx)(n− 1 )xn−^2 dx
]
=−xncosx+nxn−^1 (sinx)
−n(n− 1 )
∫
xn−^2 sinxdx
i.e. In=−xncosx+nxn−^1 sinx−n(n−1)In− 2 (3)