Integration techniques: substitution rule
Substitution for definite integrals
For nice functions f and g ,
Z
b
0
Z
g (b)
f (g (x)) g (x) dx =
a
Math 105 (Section 204)
f (u) du.
g (a)
Integration techniques
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Integration techniques: substitution rule
Substitution for definite integrals
For nice functions f and g ,
Z
b
0
Z
g (b)
f (g (x)) g (x) dx =
a
f (u) du.
g (a)
Remarks:
Do not forget to change limits after substitution!
Typically, subsitution should be used to simplify the integral, so that
the right hand side is in a form that can be computed directly.
Math 105 (Section 204)
Integration techniques
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Substitution rule: an application
Use an appropriate substitution to evaluate the following integral:
ln 2
Z
0
ex
dx.
1 + e 2x
A. ln 5
B. 2/5
C. arctan(5)
D. arctan(2) −
π
4
Math 105 (Section 204)
Integration techniques
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Integration techniques : integration by parts
Recall the product rule for differentiation: if u and v are two differentiable
functions, then
0
(u(x)v (x)) = u 0 (x)v (x) + u(x)v 0 (x).
In other words, the antiderivative of u 0 v + uv 0 is uv , i.e.,
Z
0
u (x)v (x) + u(x)v 0 (x) dx = u(x)v (x) + C .
Integration by parts formula
Z
0
Z
u (v )v (x) dx = u(x)v (x) −
Math 105 (Section 204)
Integration techniques
u(x)v 0 (x) dx + C .
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How to recognize an integral requiring integration by parts?
The integrand has to be a product of two functions, one of which is
easy to integrate (namely u 0 ), the other easy to differentiate (namely
v ).
Math 105 (Section 204)
Integration techniques
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How to recognize an integral requiring integration by parts?
The integrand has to be a product of two functions, one of which is
easy to integrate (namely u 0 ), the other easy to differentiate (namely
v ).
Often the integrand is a product of two functions that are
fundamentally dissimilar, such as xe x or e x sin x or x arctan x.
Math 105 (Section 204)
Integration techniques
2011W T2
4/4
How to recognize an integral requiring integration by parts?
The integrand has to be a product of two functions, one of which is
easy to integrate (namely u 0 ), the other easy to differentiate (namely
v ).
Often the integrand is a product of two functions that are
fundamentally dissimilar, such as xe x or e x sin x or x arctan x.
Keep in mind that u 0 could be 1! Sometimes the integrand is a single
function f (x) such as arcsin x or arctan x or ln x. One knows the
derivative for such f . In such situations set u 0 = 1, v = f .
Math 105 (Section 204)
Integration techniques
2011W T2
4/4
How to recognize an integral requiring integration by parts?
The integrand has to be a product of two functions, one of which is
easy to integrate (namely u 0 ), the other easy to differentiate (namely
v ).
Often the integrand is a product of two functions that are
fundamentally dissimilar, such as xe x or e x sin x or x arctan x.
Keep in mind that u 0 could be 1! Sometimes the integrand is a single
function f (x) such as arcsin x or arctan x or ln x. One knows the
derivative for such f . In such situations set u 0 = 1, v = f .
Application of integration by parts should either simplify the
integrand, or result in an integrand of the same complexity. In the
last case, a second “careful” application of integration by parts often
results in an identity involving the original integral, which can then be
manipulated to find the antiderivative (see example).
Math 105 (Section 204)
Integration techniques
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Integration by parts: an application
Evaluate the definite integral
Z
1
x 2 arcsin x dx.
0
A. 5/6
B. π/2
C. 2/9
D. 1/3
Math 105 (Section 204)
Integration techniques
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