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Chapter 6: Techniques of Integration
Section 6.6: Rationalizing Substitutions
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Example 6.6.3
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Evaluate the indefinite integral .
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Solution
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Mathematical Solution
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The integrand is real, except at where the denominator is zero. An antiderivative that is real for all such values of is
The rationalizing substitution can be used to obtain this antiderivative.
After completing the square in the denominator and applying a partial fraction decomposition, the antiderivative is given in terms of logarithms. Alternatively, after completing the square in the denominator, an antiderivative of from Table 3.10.1 is invoked.
Consequently, the alternative antiderivative is
which is real except at those points where it is undefined.
Figure 6.6.3(a) is a graph of and the line (in green). Where , the graph is drawn in black; where , it is drawn in red.
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f:=proc(s)
local C,CC;
C := (3/13)*sqrt(13)*abs(tan((1/2)*x)-2/3);
CC:=evalf(eval(C,x=s));
if CC<1 then CC else 10;end if;
end proc:
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g:=proc(s)
local C,CC;
C := (3/13)*sqrt(13)*abs(tan((1/2)*x)-2/3);
CC:=evalf(eval(C,x=s));
if CC>1 then CC else -10;end if;
end proc:
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p1:=plot(f,-9..9,0..3,discont=true,color=black):
p2:=plot(g,-9..9,0..3,discont=true,color=red):
p3:=plot(1,x=-9..9,color=green):
plots:-display(p1,p2,p3,scaling=constrained);
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Figure 6.6.3(a) in black; in red
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Maple Solution
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Maple evaluates the given integral in terms of the inverse hyperbolic tangent, a function that is real only when the absolute value of its argument is less than 1.
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Evaluate the integral
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Control-drag the given integral.
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Context Panel: Evaluate and Display Inline
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=
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Table 6.6.3(a) shows the first few steps in the annotated stepwise solution generated by the
tutor when the Constant, Constant Multiple, and Sum rules are taken as Understood Rules.
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Table 6.6.3(a) Initial steps taken by the Integration Methods tutor
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Maple applies the rationalizing substitution and imposes a partial-fraction decomposition. Hence, the antiderivative will contain two logarithms, but Maple will not take the absolute value of the arguments. In fact, the ultimate solution generated by the tutor is
an expression that is precisely equivalent to Maple's solution in terms of the inverse hyperbolic tangent.
Table 6.6.3(b) lists the Maple calculations that would result in the solution given in terms of logarithms of absolute values.
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Initialize
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Install the IntegrationTools package.
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Assign the inert form of the integral the name .
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Apply the rationalizing substitution
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Partial-fraction decomposition of the integrand
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Integrate the partial fractions
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Use "brute-force" to take the absolute value of the argument of the logarithms
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Table 6.6.3(b) Stepwise calculation of a solution given in terms of logarithms of absolute values.
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Note that an annotated stepwise solution is available via the Context Panel with the "All Solution Steps" option.
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The rules of integration can also be applied via the Context Panel, as per the figure to the right.
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