Understanding Engineering Mathematics

Or: ∫

xsin(x^2 + 1 )dx=

1

2

∫

sin(x^2 + 1 )d(x^2 + 1 )

=−

1

2

cos(x^2 + 1 )+C

(iii) Now you may be able to appreciate the following more easily:

∫

cosxesinxdx=

∫

esinx(cosxdx)

=

∫

esinxd(sinx)

=esinx+C

(iv)

∫

sinxcosxdx=

∫

sinxd(sinx)=

1

2

sin^2 x+C

Compare this with the result that you might obtain for Q9.1.8(iv). In

that case you may obtain

−^14 cos 2x+C

Using the double angle formula for cos 2x( 188

➤

)shows that this

in fact only differs from the answer obtained above by a constant,

which is unimportant because both forms of the answer contain an

arbitrary constant.

9.2.7 Integrating rational functions

➤

252 283➤

Any rational function has the form:

polynomial

polynomial

We can assume that the numerator has lower degree than the denominator, otherwise
we could divide out to get a polynomial plus such a fraction. In many useful cases the
polynomial in the denominator can be factorised into linear and/or quadratic factors with
real coefficients. We could then use partial fractions and substitution to split the rational
function up into integrals of the general form

∫

ax+b

cx^2 +dx+e

dx

This form of integral is therefore very important, and we will study it in detail. How we
approach it depends on whether or not the denominator factorises (45
➤
). If it does, we
can usepartial fractions(62
➤
). If it doesn’t thencompleting the square(66
➤
) enables
us to use a linear substitution and a standard integral.