2.2 INTEGRATION
Evaluate the integralI=∫
1
x^2 +4x+7dx.We can write the integral in the form
I=∫
1
(x+2)^2 +3dx.Substitutingy=x+ 2, we finddy=dxand hence
I=∫
1
y^2 +3dy,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 partsIntegration 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)=udv
dx+du
dxv,whereuandvare functions ofx, we now integrate to find
uv=∫
udv
dxdx+∫
du
dxvdx.Rearranging into the standard form for integration by parts gives
∫
u
dv
dxdx=uv−∫
du
dxvdx. (2.36)Integration by parts is often remembered for practical purposes in the formthe 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.