426 INTEGRAL CALCULUS

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−6 cost

Thus
∫ 2

1

4 t^3 costdt

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

=[4(8 sin 2+12 cos 2−12 sin 2−6 cos 2)]

−[4(sin 1+3 cos 1−6 sin 1−6 cos 1)]

=(− 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)