Honors AP Calculus BC Trig Integration Techniques 13 December 2013

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1 Honors AP Calculus BC Name: Trig Integration Techniques 13 December 2013 Integration Techniques Antidifferentiation Substitutiion (antidifferentiation of the Chain rule) Integration by Parts (antidifferentiation of the Product rule) o straightforward o repeated (tabular) o circular Trigonometric Substitution (for a 2 x 2, x 2 a 2, a 2 + x 2 forms see page 2) Separation of Variables (lesson on December 16 th ) 1 Partial Fraction Decomposition (when is involved lesson on December (x a )(x b ) 18h ) Powers of Trig Functions (see pages 3-5) Laplace Transforms (after Christmas break) Solving Differential Equations (specifically Initial Value Problems) Graphically (Slope Fields draw possible solution from initial point) Numerically (Euler s Method calculate approximate values from initial point) Analytically (using integration techniques, initial value determines C) Analytically, we can currently solve differential equations of the form: = ax and = af (x) [antidifferentiation] dx dx dx = a f ʹ (g(x)) g ʹ (x) [substitution] = af (x) g ʹ (x) [integration by parts] dx but we cannot yet solve differential equations of the form: = ay and = af (x) g(y) [separation of variables] dx dx = f (x) + g(y) dx d 2 y dx = a y ʹ + by (more often written as f (x) = y ʹ + 2by ʹ + cy) 2 The last two forms require a technique called Laplace Transforms which is not in the book but is really interesting and useful.

2 Trig Integration Techniques page 2 Trig Substitution In finding the area of a circle, we usually employ a geometric method rather than trying to integrate r 2 x 2 dx. However, this form shows up in situations where a purely geometric method is not available. In this case we will use another form of substitution called trigonometric substitution. Trigonometric Substitution is based on the relationship of the sides in a right triangle. We are able to use another variable to simplify the integral. sinθ = x a x = a sinθ tanθ = x a x = a tanθ secθ = x a x = a secθ a 2 x 2 = a 2 (asinθ) 2 = a 2 (1 sin 2 θ) = acosθ a 2 + x 2 = a 2 + (atanθ) 2 = a 2 (1+ tan 2 θ) = asecθ x 2 a 2 = (asecθ) 2 a 2 = a 2 (sec 2 θ 1) = atanθ Let us look at the example of the circle: 9 x 2 dx. In this case, 9 x 2 is a 2 x 2 so a = 3, x = 3sinθ and 9 x 2 = 3cosθ. We also need to keep in mind that dx = 3cosθ dθ. 9 x 2 dx = 3cosθ 3cosθdθ = 9 cos 2 θdθ = 9 ( 1 cos(2θ ) dθ = 9 1 2θ 1 sin(2θ) 4 Using x = 3sinθ to solve for θ and substituting back in gives9 1 2 sin 1 x 3 Not every problem has such a messy answer. + C ( 1 4 sin( 2sin 1 ( x 3) )) + C. 9 x 2 dx looks worse as a problem but see what happens when we do the trigonometric substitution used in the last example. 9 x 2 dx = x 2 3cosθ 3cosθdθ = cot 2 θdθ = csc 2 θ 1 9sin 2 θ x 2 + C = cot sin 1 x 3 dθ = cotθ θ ( ) sin 1 ( x 3) + C Problems in the book: p.338, 340/47 52 and 81 84

3 Trig Integration Techniques page 3 Powers of Trig Functions We have done the following trigonometric integration problems: sin(bx)dx = 1 b cos(bx) + C cos(bx)dx = 1 b sin(bx) + C (use substitution with u = sine or cosine) sin(bx)cos(bx)dx = 1 2b sin 2 (bx) + C = 1 2b cos 2 (bx) + C sec 2 (bx)dx = 1 b tan(bx) + C sec(bx)tan(bx)dx = 1 b sec(bx) + C (rewrite tan(bx) in terms of sine and cosine, then use substitution) tan(bx)dx = 1 b lncos(bx) + C = 1 b lnsec(bx) + C We have also dealt with sin 2 (bx)dx and cos 2 (bx)dx by using the power reducing identities derived last year: sin 2 x = 1 ( 2 1 cos(2x) ) and cos2 x = 1 ( 2 1+ cos(2x) ). However, we have struggled a bit when the powers on the trig functions got higher, for example sec 3 (bx)dx. Below are examples of techniques available to address such problems. All of the examples have an integrand with sin 3 (x) and an increasing number of cosines. Example 1: sin 3 (x)dx sin 3 (x)dx = 1 cos 2 x sin(x)dx = sin(x)dx u 2 du Example 2: sin 3 (x)cos(x)dx sin 3 (x)cos(x)dx = u 3 du Example 3: sin 3 (x)cos 2 (x)dx sin 3 (x)cos 2 (x)dx = sin(x) ( 1 cos 2 x)cos 2 (x)dx = ( cos 2 x cos 4 x)sin(x)dx = ( u 4 u 2 )du Example 4: sin 3 (x)cos 3 (x)dx sin 3 (x)cos 3 (x)dx = sin(x) ( 1 cos 2 x)cos 3 (x)dx and finish like example 3 or sin 3 (x)cos 3 (x)dx = ( sin(x)cos(x) ) 3 1 dx = sin(2x) 3 ( 2 ) dx and finish like example 1. These examples lead us to some strategies for evaluating sin m (x)cos n (x)dx (a) If the power of the sine is odd (m is odd), save one sine factor and use sin 2 (x) = 1 cos 2 (x) to express the remaining factors in terms of cosine (b) If the power of the cosine is odd (n is odd), save one cosine factor and use cos 2 (x) = 1 sin 2 (x) to express the remaining factors in terms of sine (c) If the powers are both odd, either technique can be used or the half-angle identity sin(x)cos(x) = 1 2 sin(2x) can be used (see example 4 above). (d) If the powers of both sine and cosine are even, use the power-reducing identities mentioned above and then revisit strategies a, b, or c.

4 Trig Integration Techniques page 4 There are similar examples and strategies when working with tangent and secant but we need to be able to find the indefinite integral sec(x)dx first. sec(x) + tan(x) sec(x)dx = sec(x) sec(x) + tan(x) dx = sec 2 (x) + sec(x)tan(x) dx. sec(x) + tan(x) At this point we substitute u = sec(x) + tan(x) and du = (sec(x)tan(x) + sec 2 (x))dx. so du sec(x)dx = = lnsec(x) + tan(x) + C u Example 5: tan 3 (x)dx tan 3 (x)dx = tan(x) sec 2 x 1 dx = u du tan(x)dx Example 6: tan 3 (x)sec(x)dx tan 3 (x)sec(x)dx = tan 2 (x)tan(x)sec(x)dx = ( sec 2 x 1)tan(x)sec(x)dx = u 2 du tan(x)sec(x)dx Example 7: tan 3 (x)sec 2 (x)dx tan 3 (x)sec 2 (x)dx = u 3 du Example 8: tan 3 (x)sec 3 (x)dx tan 3 (x)sec 3 (x)dx = ( sec 2 x 1)sec 2 (x) tan(x)sec(x)dx u 2 du = u 2 1 Example 9: sec 3 (x)dx Your first thought might be to rewrite this as ( tan 2 (x) +1)sec(x)dx but this will lead to a circular situation that ends with 0 = 0. It is better to use integration by parts with u = sec(x) and dv = sec 2 (x)dx. sec 3 (x)dx = sec(x)tan(x) sec(x)tan 2 (x)dx = sec(x)tan(x) sec(x) ( sec 2 (x) 1)dx = sec(x)tan(x) sec 3 (x)dx + sec(x)dx Strategies for evaluating tan m (x)sec n (x)dx (a) If the power of tangent is odd (m is odd), save a factor of sec(x)tan(x) and use tan 2 (x) = sec 2 (x) 1 to express the remaining factors in terms of secant. (b) If the power of secant is even (n is even), save a factor of sec 2 (x) and use sec 2 (x) = 1 + tan 2 (x) to express the remaining factors in terms of tangent. (c) Other cases are not as clear cut and will require trying strategies like integration by parts with the identities.

5 Trig Integration Techniques page 5 Final notes: (a) cosecant and cotangent follow similar strategies to secant and tangent. Be careful with the negatives. csc 2 (x)dx = cot(x) + C csc(x)cot(x)dx = csc(x) + C cot(x)dx = lnsin(x) + C = lncsc(x) + C (b) If you must evaluate sin(mx)cos(nx)dx then it is helpful to employ the identities: sin AcosB = 1 [ 2 sin(a B) + sin(a + B) ] sin AsinB = 1 [ 2 cos(a B) cos(a + B) ] cos AcosB = 1 [ 2 cos(a B) + cos(a + B) ] Problems 1. cos 3 (x)dx 2. sin 5 (x)cos 2 (x)dx 3. sin 4 (x)dx 4. sin 2 (πx)cos 4 (πx)dx 5. tan 6 (x)sec 4 (x)dx 6. tan 3 (2x)sec 5 (2x)dx 7. cot 3 (x)csc 3 (x)dx 8. cot 4 (x)csc 6 (x)dx 9. sin(8x)cos(5x)dx 10. cos(πx)cos(4πx)dx