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Techniques of Integration

الكلية كلية الهندسة     القسم  الهندسة الميكانيكية     المرحلة 1
أستاذ المادة احمد كاظم حسين الحميري       15/12/2016 16:26:11
The rules of differentiation give us an explicit algorithm for calculating derivatives of all elementary
functions, including trigonometric and exponential functions, as well as logarithms. By
comparison, integration of elementary functions in general is a more difficult task. In fact, some
integrals, such as
Z e?x2
dx, Z sin(x2) dx, Z sin x
x
dx, Z dx
ln x
cannot be expressed as elementary functions. To understand better this striking phenomenon,
recall that R dx
x = ln x, and we see that integration of a rational function leads to a transcendental
function. So one may expect that integration of transcendental functions leads to an even bigger
class of functions that cannot be expressed as combinations of elementary functions, although from
the general theory of integration we know that these functions exist and are well-defined.
In this lecture we discuss some more advanced techniques of integration.
2.1. Integration by Parts. Let u = f(x) and v = g(x) be differentiable functions with continuous
derivatives. Then by the product rule: d(uv) = udv + vdu, or
(2.1) udv = d(uv) ? vdu.
The antiderivative of the expression d (uv) is uv, and therefore, by integrating both sides of (2.1)
we obtain the formula of integration by parts:
(2.2) Z u dv = uv ? Z v du.
The integration by parts formula can be used, for example, for integration of products of functions.
Example 2.1. To evaluate R x cos xdx we let u = x, dv = cos xdx, so that du = dx and v = sin x,
and use (2.2):
(2.3) Z x cos xdx = x sin x ? Z sinxdx = x sin x + cos x + C.
?
Application of the integration by parts formula requires breaking the integrand into two parts:
u and dv, the first of which should be differentiated and the second integrated. The choice of u
and dv should be such that integration of dv is relatively simple and that the the resulting integral
is simpler than the original.
Example 2.2. To compute R x3 ln x dx we set u = ln x, dv = x3dx. Then
Z x3 ln x dx =
1
4
x4 ln x ?
1
4 Z x3dx =
1
4
x4 ln x ?
1
16
x4 + C.
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