2.2 INTEGRATION
Evaluate the integral
I=
∫
1
x^2 +4x+7
dx.
We can write the integral in the form
I=
∫
1
(x+2)^2 +3
dx.
Substitutingy=x+ 2, we finddy=dxand hence
I=
∫
1
y^2 +3
dy,
Hence, by comparison with the table ofstandard integrals (see subsection 2.2.3)
I=
√
3
3
tan−^1
(
y
√
3
)
+c=
√
3
3
tan−^1
(
x+2
√
3
)
+c.
2.2.8 Integration by parts
Integration by parts is the integration analogy of product differentiation. The
principle is to break down a complicated function into two functions, at least one
of which can be integrated by inspection. The method in fact relies on the result
for the differentiation of a product. Recalling from (2.6) that
d
dx
(uv)=u
dv
dx
+
du
dx
v,
whereuandvare functions ofx, we now integrate to find
uv=
∫
u
dv
dx
dx+
∫
du
dx
vdx.
Rearranging into the standard form for integration by parts gives
∫
u
dv
dx
dx=uv−
∫
du
dx
vdx. (2.36)
Integration by parts is often remembered for practical purposes in the form
the integral of a product of two functions is equal to{the first times the integral of
the second}minus the integral of{the derivative of the first times the integral of
the second}. Here,uis ‘the first’ anddv/dxis ‘the second’; clearly the integralv
of ‘the second’ must be determinable by inspection.
Evaluate the integralI=
∫
xsinxdx.
In the notation given above, we identifyxwithuand sinxwithdv/dx. Hencev=−cosx
anddu/dx= 1 and so using (2.36)
I=x(−cosx)−
∫
(1)(−cosx)dx=−xcosx+sinx+c.