8 TECHNIQUES OF INTEGRATION TECHNIQUES OF INTEGRATION 7

8 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 would be effective. § However, as it stands, , the substitution is more difficult.

TRIGONOMETRIC SUBSTITUTION If we change the variable from x to θ by the substitution x = a sin θ, the identity 1 – sin 2θ = cos 2θ lets us lose the root sign. § This is because:

TRIGONOMETRIC SUBSTITUTION Notice the difference between the substitution u = a 2 – x 2 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 g has 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 θ to 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 θ dθ 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 θ in 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 θ from the figure: § Although θ > 0 here, this expression for cot θ is 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 To evaluate this integral, we substitute x = a sin θ. Then, dx = a cos θ dθ. Example 2

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 Also, since 0 ≤ θ ≤ π/2, Example 2

TRIGONOMETRIC SUBSTITUTION Therefore, Example 2

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 πr 2.

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 Find § Let x = 2 tan θ, –π/2 < θ < π/2. § Then, dx = 2 sec 2 θ dθ and Example 3

TRIGONOMETRIC SUBSTITUTION § Thus, we have: Example 3

TRIGONOMETRIC SUBSTITUTION Example 3 To evaluate this trigonometric integral, we put everything in terms of sin θ and cos θ:

TRIGONOMETRIC SUBSTITUTION Example 3 zz Therefore, making the substitution u = sin θ, we have:

TRIGONOMETRIC SUBSTITUTION We use the figure to determine that: § Hence, Example 3

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 = x 2 + 4 is simpler. § This is because, then, du = 2 x dx and

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 θ tan θ dθ and

TRIGONOMETRIC SUBSTITUTION Therefore, E. g. 5—Solution 1

TRIGONOMETRIC SUBSTITUTION E. g. 5—Solution 1 The triangle in the figure gives:

TRIGONOMETRIC SUBSTITUTION So, we have: E. g. 5—Solution 1

TRIGONOMETRIC SUBSTITUTION E. g. 5—Sol. 1 (For. 1) Writing C 1 = 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 cosh 2 y – sinh 2 y = 1, we have:

TRIGONOMETRIC SUBSTITUTION Since dx = a sinh t dt, we obtain: E. g. 5—Solution 2

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 = 2 x.

TRIGONOMETRIC SUBSTITUTION Example 6 When we combine this with the tangent substitution, we have This gives . and

TRIGONOMETRIC SUBSTITUTION Example 6 When x = 0, tan θ = 0; so θ = 0. When x = , tan θ = ; so θ = π/3.

TRIGONOMETRIC SUBSTITUTION Example 6

TRIGONOMETRIC SUBSTITUTION Example 6 Now, we substitute u = cos θ so that du = - sin θ dθ. § When θ = 0, u = 1. § When θ = π/3, u = ½.

TRIGONOMETRIC SUBSTITUTION Therefore, Example 6

TRIGONOMETRIC SUBSTITUTION Example 7 Evaluate § We can transform the integrand into a function for which trigonometric substitution is appropriate, by first completing the square under the root sign:

TRIGONOMETRIC SUBSTITUTION Example 7 This suggests we make the substitution u = x + 1. § Then, du = dx and x = u – 1. § So,

TRIGONOMETRIC SUBSTITUTION We now substitute This gives and Example 7 .

TRIGONOMETRIC SUBSTITUTION So, Example 7

TRIGONOMETRIC SUBSTITUTION The figure shows the graphs of the integrand in Example 7 and its indefinite integral (with C = 0). § Which is which?