Higher Engineering Mathematics, Sixth Edition

434 Higher Engineering Mathematics

=x(lnx)n−n

∫

(lnx)n−^1 dx

i.e.In=x(lnx)n−nIn− 1

Whenn=3,

∫

(lnx)^3 dx=I 3 =x(lnx)^3 − 3 I 2

I 2 =x(lnx)^2 − 2 I 1 andI 1 =

∫

lnxdx=x(lnx− 1 )from

Problem 7, page 422.

Hence

∫

(lnx)^3 dx=x(lnx)^3 −3[x(lnx)^2 − 2 I 1 ]+c

=x(lnx)^3 −3[x(lnx)^2

−2[x(lnx− 1 )]]+c

=x(lnx)^3 −3[x(lnx)^2

− 2 xlnx+ 2 x]+c

=x(lnx)^3 − 3 x(lnx)^2

+ 6 xlnx− 6 x+c

=x[(lnx)^3 −3(lnx)^2

+6lnx−6]+c

Now try the following exercise

Exercise 172 Further problems on

reduction formulae

1. Evaluate

∫ π 2

0

cos^2 xsin^5 xdx.

[

8

105

]

2. Determine

∫

tan^6 xdxby usingreduction for-

mulae and hence evaluate

∫ π

4

0

tan^6 xdx.

[

13

15

−

π

4

]

3. Evaluate

∫ π 2

0

cos^5 xsin^4 xdx.

[

8

315

]

4. Use a reduction formula to determine∫
   (lnx)^4 dx.
   [
   x(lnx)^4 − 4 x(lnx)^3 + 12 x(lnx)^2
   − 24 xlnx+ 24 x+c

]

5. Show that

∫ π

2

0

sin^3 θcos^4 θdθ=

2

35