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Home : Support : Online Help : Education : Study Guides : Calculus : Chapter 6 - Techniques of Integration : Examples : Section 6-6 : Example 6-6-5

Chapter 6: Techniques of Integration

Section 6.6: Rationalizing Substitutions

Example 6.6.5

Obtain a continuous antiderivative for $1 / (7 + 2 \sin (x) + 3 \cos (x))$ .

Solution

Mathematical Solution

The following implementation of the rationalizing substitution $z = \tan (x / 2)$ results in a discontinuous antiderivative.

$\int \frac{dx}{7 + 2 \sin (x) + 3 \cos (x)}$	$= \int \frac{\frac{2 dz}{1 + z^{2}}}{7 + 2 (\frac{2 z}{1 + z^{2}}) + 3 (\frac{1 - z^{2}}{1 + z^{2}})}$
	$= \int \frac{dz}{2 z^{2} + 2 z + 5}$
	$= \int \frac{dz}{2 ({(z + 1 / 2)}^{2} + 9 / 4)}$
	$= \int \frac{dz}{2 \cdot \frac{9}{4} (\frac{4}{9} {(z + \frac{1}{2})}^{2} + 1)}$
	$= \frac{2}{9} \int \frac{dz}{{(\frac{2}{3} (z + \frac{1}{2}))}^{2} + 1}$
	$= \frac{2}{9} \cdot \frac{3}{2} \int \frac{du}{u^{2} + 1}$
	$= \frac{1}{3} \arctan (u)$
	$= \frac{1}{3} \arctan (\frac{2}{3} (z + 1 / 2))$
	$= \frac{1}{3} \arctan ((2 \tan (x / 2) + 1) / 3)$

Table 6.6.5(a) lists a relevant integration formula taken from a standard table of integrals.

$\int \frac{dx}{a + b \sin (x) + c \cos (x)}$ = ${\begin{array}{c} \frac{1}{u} \ln (\frac{b - u + (a - c) \tan (x / 2}{b + u + (a - c) \tan (x / 2)}) & a^{2} < b^{2} + c^{2} \\ \frac{2}{v} \arctan (\frac{b + (a - c) \tan (x / 2)}{v}) & a^{2} > b^{2} + c^{2} \end{array}$

$u = \sqrt{b^{2} + c^{2} - a^{2}}, v = \sqrt{a^{2} - b^{2} - c^{2}}$

Table 6.6.5(a) Relevant integration formula from a table of integrals

With $a = 7, b = 2, c = 3$ , the second rule in Table 6.6.5(a) applies because $a^{2} = 49 > 4 + 9 = 13 = b^{2} + c^{2}$ . Hence, $v = \sqrt{49 - 13} = \sqrt{36} = 6$ , and

$\int \frac{dx}{7 + 2 \sin (x) + 3 \cos (x)}$	$= \frac{2}{6} \arctan (\frac{2 + (7 - 3) \tan (x / 2)}{6})$
	$= \frac{1}{3} \arctan (\frac{1}{3} (1 + 2 \tan (x / 2)))$

Figure 6.6.5(a) shows that this antiderivative is discontinuous. In fact, jumps of $π / 3$ occur at $x = (2 k + 1) π$ for any integer $k$ . If, between the discontinuities an appropriate multiple of $π / 3$ can be added to the integrand, the result would be the continuous antiderivative shown in Figure 6.6.5(b).

Figure 6.6.5(a) Discontinuous antiderivative

module()
local p1,p2,p3,L,R;
L:=(1/3)*arctan((2/3)*tan((1/2)*x)+1/3)-(1/3)*arctan(1/3)-piecewise(-(1/2)*(Pi-x)/Pi < -1, -(1/3)*ceil(-(1/2)*(Pi-x)/Pi)*Pi, 0);
R:=-(1/3)*arctan(1/3)+(1/3)*arctan((2/3)*tan((1/2)*x)+1/3)+piecewise(0 < -Pi+x, (1/3*(floor(-(1/2)*(Pi-x)/Pi)+1))*Pi, 0);
p1:=plot(L,x=-3*Pi..0);
p2:=plot(R,x=0..3*Pi);
p3:=plots:-display(p1,p2);
print(p3);
end module:

Figure 6.6.5(b) Continuous antiderivative

Maple Solution

Obtain the discontinuous antiderivative and determine the jump at the discontinuities.

Evaluate the integral and name the antiderivative $g$

$\int \frac{1}{7 + 2 \sin (x) + 3 \cos (x)} &DifferentialD; x$ = $\frac{1}{3} \arctan (\frac{1}{3} + \frac{2}{3} \tan (\frac{1}{2} x))$ $\overset{assign to a name}{\to}$ $g$

Calculate the jump at $x = π$

•	Expression palette: Limit template

•	Calculate the difference in the left and right limits.

$lim_{x \to π -} g - lim_{x \to π +} g$ = $\frac{1}{3} π$

Obtain a continuous antiderivative by evaluating an appropriate definite integral.

•	Expression palette: Definite Integral template

•	Context Panel: 2-D Math≻Convert To≻Inert Form

•	Press the Enter key.

$Q ≔ \int_{0}^{x} \frac{1}{7 + 2 \sin (t) + 3 \cos (t)} &DifferentialD; t &colon;$

•	Evaluate the definite integral under the separate assumptions that $x$ is positive and $x$ is negative.

$simplify (value (Q)) assuming x > 0$

${\begin{array}{c} - \frac{1}{3} \arctan (\frac{1}{3}) + \frac{1}{3} \arctan (\frac{1}{3} + \frac{2}{3} \tan (\frac{1}{2} x)) & x \leq π \\ \frac{1}{3} π floor (- \frac{1}{2} \frac{π - x}{π}) - \frac{1}{3} \arctan (\frac{1}{3}) + \frac{1}{3} \arctan (\frac{1}{3} + \frac{2}{3} \tan (\frac{1}{2} x)) + \frac{1}{3} π & π < x \end{array}$

$simplify (value (Q)) assuming x < 0$

${\begin{array}{c} - \frac{1}{3} \arctan (\frac{1}{3}) + \frac{1}{3} \arctan (\frac{1}{3} + \frac{2}{3} \tan (\frac{1}{2} x)) + \frac{1}{3} ceil (- \frac{1}{2} \frac{π - x}{π}) π & x < - π \\ - \frac{1}{3} \arctan (\frac{1}{3}) + \frac{1}{3} \arctan (\frac{1}{3} + \frac{2}{3} \tan (\frac{1}{2} x)) & - π \leq x \end{array}$

If $g (x)$ is defined as

$g (x) = - \frac{1}{3} \arctan (\frac{1}{3}) + \frac{1}{3} \arctan (\frac{1}{3} + \frac{2}{3} \tan (\frac{1}{2} x))$

then the continuous antiderivative can be written as

${\begin{array}{c} g (x) + \frac{1}{3} ceil (\frac{x - π}{2 π}) & x < - π \\ g (x) & x \in [- π, π] \\ g (x) + \frac{π}{3} (1 + floor (\frac{x - π}{2 π})) & x > π \end{array}$

where ceil is the "greatest integer" function and floor is the "least integer" function. These two functions tack onto $g (x)$ the appropriate multiple of $π / 3$ to counterbalance the jumps in the discontinuous $g (x)$ .

This example concludes with Table 6.6.5(b) in which the antiderivative $\frac{1}{3} \arctan (\frac{1}{3} (1 + 2 \tan (x / 2)))$ is obtained from first principles is obtained in Maple.

•	Expression palette: Indefinite Integral template

•	Context Panel: 2-D Math≻Convert To≻Inert Form

$Q ≔ \int \frac{1}{7 + 2 \sin (x) + 3 \cos (x)} &DifferentialD; x &colon;$

•	Apply the rationalizing substitution $z = \tan (x / 2)$ via the Change command

$q_{1} ≔ IntegrationTools :- Change (Q, z = \tan (x / 2), z)$

$\int \frac{2}{(7 + 2 \sin (2 \arctan (z)) + 3 \cos (2 \arctan (z))) (z^{2} + 1)} &DifferentialD; z$

•	Convert the integrand to a rational function.

$q_{2} ≔ simplify (expand (q_{1}))$

$\int \frac{1}{2 z^{2} + 2 z + 5} &DifferentialD; z$

•	Complete the square in the denominator. (Maple can find an antiderivative without this step, but completion of the square is an essential step in working from first principles.)

$q_{3} ≔ Student :- Precalculus :- CompleteSquare (q_{2})$

$\int \frac{1}{2 {(z + \frac{1}{2})}^{2} + \frac{9}{2}} &DifferentialD; z$

•	Evaluate the integral.

$q_{4} ≔ value (q_{3})$

$\frac{1}{3} \arctan (\frac{2}{3} z + \frac{1}{3})$

•	Revert the rationalizing substation.

$eval (q_{4}, z = \tan (x / 2))$

$\frac{1}{3} \arctan (\frac{1}{3} + \frac{2}{3} \tan (\frac{1}{2} x))$

Table 6.6.5(b) Maple calculation of antiderivative from first principles

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