(ii) There are a number of ways we might tackle sin^3 x, covered in
Section 9.2.8, but perhaps the easiest is to use the Pythagorean
identity to write sin^2 x= 1 −cos^2 xand then use the substitution,
u=cosx(271➤
).(iii) Inelnx
xyou may notice that1
xis the derivative of lnx, suggesting
the substitutionu=lnx. But hopefully you are now happy enough
with the exponential function to note the simplificationelnx=x
and take it from there (259➤
)!(iv) Inx− 1
2 x^2 +x− 3the top is not the derivative of the bottom, but we
notice that the denominator factorises into( 2 x+ 3 )(x− 1 )and so
the function can be simplified by cancelling thex−1 (providedx
is not equal to 1, of course), resulting in a simple linear substitution
and log integral (263
➤
).(v) No,xe^3 x
2
is not a prime candidate for integration by parts, despite
being similar to products we have dealt with in that way. Instead, we
notice that thexis almost the derivative of thex^2 in the exponent,
and this suggests that the substitutionu=x^2 will rescue us here
(263
➤
).(vi) Nasty looking functions such as3
√
3 − 4 x−x^2containing roots of
algebraic functions usually succumb only to some sort of trig or
hyperbolic substitution as considered in Section 9.2.9 – though in
this case we need to complete the square first. A similar integral is
done in Review Question 9.1.9(ii).
(vii)xsin(x+ 1 ) yields to a straightforward integration by parts, in
which the sin(x+ 1 )is integrated first (273➤
).(viii) cos^4 xcan be integrated by using the double angle formulae
cos 2x=2cos^2 x− 1 = 1 −2sin^2 x
to replace cos^2 xand subsequently cos^22 xto leave us with integrals
of a constant, cos 2xand cos 4x(270➤
).(ix) By now you should see immediately that ln(
ex
x)
=x−lnxand
we have only to integrate the lnx, which can be done by parts, as
shown in Section 9.2.10.(x) The denominator ofx+ 2
x^2 − 5 x+ 6factorises to(x− 2 )(x− 3 )and
we have a straightforward partial fractions to deal with (266➤
).
(xi)xe^2 xprovides a very typical integration by parts (273➤
).
(xii) lnexcos(ex)we note that the derivative ofexis of courseexwhich
tells us to substituteu=ex(263➤
).