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## TECHNIQUES OF INTEGRATION

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**7**TECHNIQUES OF INTEGRATION**TECHNIQUES OF INTEGRATION**• 7.3 • Trigonometric Substitution • In this section, we will learn about: • The various types of trigonometric substitutions.**TRIGONOMETRIC SUBSTITUTION**• In finding the area of a circle or an ellipse, an integral of the form arises, where a > 0. • If it were , the substitution would be effective. • However, as it stands, is more difficult.**TRIGONOMETRIC SUBSTITUTION**• If we change the variable from x to θby the substitution x = a sin θ, the identity 1 – sin2θ = cos2θlets us lose the root sign. • This is because:**TRIGONOMETRIC SUBSTITUTION**• Notice the difference between the substitution u = a2 – x2 and the substitution x = a sin θ. • In the first, the new variable is a function of the old one. • In the second, the old variable is a function of the new one.**TRIGONOMETRIC SUBSTITUTION**• In general, we can make a substitution of the form x = g(t) by using the Substitution Rule in reverse. • To make our calculations simpler, we assume ghas an inverse function, that is, g is one-to-one.**INVERSE SUBSTITUTION**• Here, if we replace u by x and x by t in the Substitution Rule (Equation 4 in Section 5.5), we obtain: • This kind of substitution is called inverse substitution.**INVERSE SUBSTITUTION**• We can make the inverse substitution x = a sin θ, provided that it defines a one-to-one function. • This can be accomplished by restricting θto lie in the interval [-π/2, π/2].**TABLE OF TRIGONOMETRIC SUBSTITUTIONS**• Here, we list trigonometric substitutions that are effective for the given radical expressions because of the specified trigonometric identities.**TABLE OF TRIGONOMETRIC SUBSTITUTIONS**• In each case, the restriction on θ is imposed to ensure that the function that defines the substitution is one-to-one. • These are the same intervals used in Section 1.6 in defining the inverse functions.**TRIGONOMETRIC SUBSTITUTION**Example 1 • Evaluate • Let x = 3 sin θ, where –π/2 ≤ θ ≤ π/2. • Then, dx = 3 cos θdθ and • Note that cos θ≥ 0 because –π/2 ≤ θ ≤ π/2.)**TRIGONOMETRIC SUBSTITUTION**Example 1 • Thus, the Inverse Substitution Rule gives:**TRIGONOMETRIC SUBSTITUTION**Example 1 • As this is an indefinite integral, we must return to the original variable x. • This can be done in either of two ways.**TRIGONOMETRIC SUBSTITUTION**Example 1 • One, we can use trigonometric identities to express cot θin terms of sin θ = x/3.**TRIGONOMETRIC SUBSTITUTION**Example 1 • Two, we can draw a diagram, where θ is interpreted as an angle of a right triangle.**TRIGONOMETRIC SUBSTITUTION**Example 1 • Since sin θ = x/3, we label the opposite side and the hypotenuse as having lengths x and 3.**TRIGONOMETRIC SUBSTITUTION**Example 1 • Then, the Pythagorean Theorem gives the length of the adjacent side as:**TRIGONOMETRIC SUBSTITUTION**Example 1 • So, we can simply read the value of cot θfrom the figure: • Although θ > 0 here, this expression for cot θis valid even when θ < 0.**TRIGONOMETRIC SUBSTITUTION**Example 1 • As sin θ= x/3, we have θ = sin-1(x/3). • Hence,**TRIGONOMETRIC SUBSTITUTION**Example 2 • Find the area enclosed by the ellipse**TRIGONOMETRIC SUBSTITUTION**Example 2 • Solving the equation of the ellipse for y, we get • or**TRIGONOMETRIC SUBSTITUTION**Example 2 • As the ellipse is symmetric with respect to both axes, the total area A is four times the area in the first quadrant.**TRIGONOMETRIC SUBSTITUTION**Example 2 • The part of the ellipse in the first quadrant is given by the function • Hence,**TRIGONOMETRIC SUBSTITUTION**Example 2 • To evaluate this integral, we substitute x = a sin θ. • Then, dx = a cos θ dθ.**TRIGONOMETRIC SUBSTITUTION**Example 2 • To change the limits of integration, we note that: • When x = 0, sin θ = 0; so θ = 0 • When x = a, sin θ = 1; so θ = π/2**TRIGONOMETRIC SUBSTITUTION**Example 2 • Also, since 0 ≤ θ ≤ π/2,**TRIGONOMETRIC SUBSTITUTION**Example 2 • Therefore,**TRIGONOMETRIC SUBSTITUTION**Example 2 • We have shown that the area of an ellipse with semiaxes a and b is πab. • In particular, taking a = b = r, we have proved the famous formula that the area of a circle with radius r is πr2.**TRIGONOMETRIC SUBSTITUTION**Note • The integral in Example 2 was a definite integral. • So, we changed the limits of integration, and did not have to convert back to the original variable x.**TRIGONOMETRIC SUBSTITUTION**Example 3 • Find • Let x = 2 tan θ, –π/2 < θ < π/2. • Then, dx = 2 sec2θdθ and**TRIGONOMETRIC SUBSTITUTION**Example 3 • Thus, we have:**TRIGONOMETRIC SUBSTITUTION**Example 3 • To evaluate this trigonometric integral, we put everything in terms of sin θand cos θ:**TRIGONOMETRIC SUBSTITUTION**Example 3zz • Therefore, making the substitution u = sin θ, we have:**TRIGONOMETRIC SUBSTITUTION**Example 3 • We use the figure to determine that: • Hence,**TRIGONOMETRIC SUBSTITUTION**Example 4 • Find • It would be possible to use the trigonometric substitution x = 2 tan θ (as in Example 3).**TRIGONOMETRIC SUBSTITUTION**Example 4 • However, the direct substitution u = x2 + 4 is simpler. • This is because, then, du = 2xdxand**TRIGONOMETRIC SUBSTITUTION**Note • Example 4 illustrates the fact that, even when trigonometric substitutions are possible, they may not give the easiest solution. • You should look for a simpler method first.**TRIGONOMETRIC SUBSTITUTION**Example 5 • Evaluate where a > 0.**TRIGONOMETRIC SUBSTITUTION**E. g. 5—Solution 1 • We let x = a sec θ, where 0 < θ < π/2 or π < θ < π/2. • Then, dx = a sec θtanθdθ and**TRIGONOMETRIC SUBSTITUTION**E. g. 5—Solution 1 • Therefore,**TRIGONOMETRIC SUBSTITUTION**E. g. 5—Solution 1 • The triangle in the figure gives:**TRIGONOMETRIC SUBSTITUTION**E. g. 5—Solution 1 • So, we have:**TRIGONOMETRIC SUBSTITUTION**E. g. 5—Sol. 1 (For. 1) • Writing C1 = C – ln a, we have:**TRIGONOMETRIC SUBSTITUTION**E. g. 5—Solution 2 • For x > 0, the hyperbolic substitution x = a cosh t can also be used. • Using the identity cosh2y – sinh2y = 1, we have:**TRIGONOMETRIC SUBSTITUTION**E. g. 5—Solution 2 • Since dx = a sinh tdt,we obtain:**TRIGONOMETRIC SUBSTITUTION**E. g. 5—Sol. 2 (For. 2) • Since cosh t = x/a, we have t = cosh-1(x/a) and**TRIGONOMETRIC SUBSTITUTION**E. g. 5—Sol. 2 (For. 2) • Although Formulas 1 and 2 look quite different, they are actually equivalent by Formula 4 in Section 3.11**TRIGONOMETRIC SUBSTITUTION**Note • As Example 5 illustrates, hyperbolic substitutions can be used instead of trigonometric substitutions, and sometimes they lead to simpler answers. • However, we usually use trigonometric substitutions, because trigonometric identities are more familiar than hyperbolic identities.**TRIGONOMETRIC SUBSTITUTION**Example 6 • Find • First, we note that • So, trigonometric substitution is appropriate.**TRIGONOMETRIC SUBSTITUTION**Example 6 • is not quite one of the expressions in the table of trigonometric substitutions. • However, it becomes one if we make the preliminary substitution u = 2x.