文档介绍：Section
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Trigonometric Substitutions
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In the last section we saw that
1 −1
Z √ dx = sin (x) + c.
1 − x2
However, we arrived at this result as a consequence of our differentiation of the arc sine
function, not as the e of the application of some systematic approach to the evalua-
tion of integrals of this type. In this section we will explore how substitutions based on the
arc sine, arc tangent, and arc secant functions provide a systematic method for evaluating
integrals similar to this one.
Sine substitutions
To begin, consider evaluating
1
Z √ dx
1 − x2
by using the substitution u = sin−1(x). The motivation for such a substitution stems from
−π≤≤π−1
the fact that, for 2 u 2 , u = sin (x) if and only if x = sin(u). In the latter form,
we see that
dx = cos(u)du
and
1 − x2 = 1 − sin2(u) = cos2(u) = | cos(u)|. ()
p q p
≥π≤≤π
Since cos(u) 0 when 2 u 2 , () es
1 − x2 = cos(u).
p
Thus
1 cos(u) −1
Z √ dx = Z du = Z du = u + c = sin (x) + c.
1 − x2 cos(u)
Of course, there is nothing new in the result itself; it is the technique, which we may
generalize to other integrals of a similar type, which is of interest. Specifically, for an
integral with a factor of the form
a2 − x2
p
or
1
√,
a2 − x2
1
2 Trigonometric Substitutions Sect