432 INTEGRAL CALCULUS=x(lnx)n−n∫
(lnx)n−^1 dxi.e.In=x(lnx)n−nIn− 1Whenn=3,
∫
(lnx)^3 dx=I 3 =x(lnx)^3 − 3 I 2
I 2 =x(lnx)^2 − 2 I 1 and I 1 =∫
lnxdx=x(lnx−1)
from Problem 7, page 420.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]+cNow try the following exercise.Exercise 173 Further problems on reduc-
tion formulae- Evaluate
∫ π
20cos^2 xsin^5 xdx.[
8
105]- Determine
∫
tan^6 xdxby using reductionformulae and hence evaluate∫ π
40tan^6 xdx.
[
13
15−π
4]- Evaluate
∫ π
20cos^5 xsin^4 xdx.[
8
315]- 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
]- Show that
∫π
20sin^3 θcos^4 θdθ=2
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