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Math 141 Lecture 6 Greg Maloney Todor Milev University of Massachusetts Boston Spring 2015 Math 141 Lecture 6 Spring 2015 Outline 1 Trigonometric Integrals Integrating rational trigonometric integrals Ad hoc methods for trigonometric integrals Math 141 Lecture 6 Spring 2015 License to use and redistribute These lecture slides and their LATEX source code are licensed to you under the Creative Commons license CC BY 3.0. You are free to Share - to copy, distribute and transmit the work, to Remix - to adapt, change, etc., the work, to make commercial use of the work, as long as you reasonably acknowledge the original project (a notice of use freecalc is sufficient). Latest version of the .tex sources of the slides: https: //sourceforge.net/p/freecalculus/code/HEAD/tree/ Should the link be outdated/moved, search for “freecalc project”. Creative Commons license CC BY 3.0: https://creativecommons.org/licenses/by/3.0/us/ and the links therein. Math 141 Lecture 6 Spring 2015 Trigonometric Integrals Integrals of the form R Integrating rational trigonometric integrals R(cos θ, sin θ)dθ, R Let R be an arbitrary rational function in two variables (quotient of polynomials in two variables). Question Can we integrate R R(cos θ, sin θ)dθ? Yes. We will learn how in what follows. The algorithm for integration is roughly: Apply the substitution θ = 2 arctan t to transform to integral of rational function. Solve as previously studied. Math 141 Lecture 6 Spring 2015 Trigonometric Integrals Integrating rational trigonometric integrals The rationalizing substitution θ = 2 arctan t R Let R- rational function in two variables. R( cos θ, sin θ)dθ can be integrated via the substitution θ = 2 arctan t. How does this transform sin θ, cos θ? How does this transform dθ? How is t expressed via θ? sin θ = cos θ = 2 tan (arctan t) 2t = 2 1 + tan (arctan t) 1 + t2 2 1 − tan (arctan t) 1 − t2 cos(2 arctan t) = = 1 + tan2 (arctan t) 1 + t2 sin(2 arctan t) = Recall the expression of sin(2z), cos(2z) via tan z: 1 cos2 z (cos2 z + sin2 z) cos12z cos2 z − sin2 z 2 sin z cos z sin (2z) = 2 sin z cos z = cos(2z) = cos2 z − sin2 z = Math 141 cos2 z + sin2 z Lecture 6 = 2 tan z 1 + tan2 z 1 cos2 z 1 cos2 z = . 1 − tan2 z . 1 + tan2 z Spring 2015 Trigonometric Integrals Integrating rational trigonometric integrals The rationalizing substitution θ = 2 arctan t R Let R- rational function in two variables. R( cos θ, sin θ)dθ can be integrated via the substitution θ = 2 arctan t. How does this transform sin θ, cos θ? How does this transform dθ? How is t expressed via θ? sin θ = cos θ = dθ = t = 2 tan (arctan t) 2t = 2 1 + tan (arctan t) 1 + t2 2 1 − tan (arctan t) 1 − t2 cos(2 arctan t) = = 1 + tan2 (arctan t) 1 + t2 2 2d (arctan t) = dt 1 + t2 θ tan 2 sin(2 arctan t) = Theorem The substitution given above transforms integral of a rational function of t. Math 141 Lecture 6 R R(cos θ, sin θ)dθ to an Spring 2015 Trigonometric Integrals Integrating rational trigonometric integrals Example Let θ = 2 arctan t, cos θ = Z dθ 2 sin θ − cos θ + 5 (complete square) 1−t 2 , 1+t 2 sin θ = Z z= √3 5 t+ 1 3 . 2dt = 2) (1 + t 2 ) − (1−t + 5 1+t 2 Z 2dt = Z 6t 2 + 4t + 4 dt = 2 Z 3t + 2t + 2 dt = 1 1 2 1 2 Z3 t + 2t 3 + 9 − 9 + 3 dt 1 = 1 2 5 3 Z t+3 +9 1 dt = 2 5 9 3 t + 1 +1 9 Math 141 2t , 1+t 2 5 Lecture 6 2 t 22t+1 3 Spring 2015 Trigonometric Integrals Integrating rational trigonometric integrals Example Let θ = 2 arctan t, cos θ = Z dθ 2 sin θ − cos θ + 5 sin θ = Z = 1 3 Z = 3 5 = = = = Math 141 1−t 2 , 1+t 2 2t , 1+t 2 z= √3 5 t+ 1 3 . dt 5 9 1 2 9 +1 5 t+ 3 √ 5 √3 t + 1 d 3 3 5 √3 5 1 3 2 +1 t+ √ Z 5 dz 2 z +1 √5 5 arctan z + C √5 3 1 5 arctan √ t+ +C 3 √5 5 5 3 θ 1 arctan √ tan + +C 5 2 3 5 Lecture 6 Spring 2015 Trigonometric Integrals Integrating rational trigonometric integrals R The integral sec θdθ appears often in practice. A quicker solution will be shown later, but first we show the standard method. Example 1 − tan2 ( 2θ ) 1 1 − t2 , dθ = 2 dt. = 2 θ 2 1 + t 1 + t2 1 + tan 2 Z Z 1 1 2 dθ = dt 1−t 2 (1 + t 2 ) cos θ 2 Z Z 1+t 2 1 1 + part. fractions dt = dt 1−t 1+t 1 − t2 − ln |1 −t| + ln |1 + t| + C 1 + t +C ln 1 − t 1 + tan θ 2 + C ln 1 − tan 2θ Set θ = 2 arctan t, cos θ = Z sec θdθ = = = = = Math 141 Lecture 6 Spring 2015 Trigonometric Integrals Integrating rational trigonometric integrals R The integral sec θdθ appears often in practice. A quicker solution will be shown later, but first we show the standard method. Example 1 − tan2 ( 2θ ) 1 1 − t2 Set θ = 2 arctan t, cos θ = , dθ = 2 dt. = 2 θ 2 2 1 + t 1 + t 1 + tan 2 Z 1 + tan θ 2 + C sec θdθ = ln 1 − tan 2θ This is a perfectly good answer, however there’s a simplification: 2 sin 2θ cos 2θ + sin2 2θ + cos2 2θ sin θ + 1 tan θ + sec θ = = cos θ cos2 2θ − sin2 2θ 2 sin 2θ + cos 2θ sin 2θ + cos 2θ = = cos 2θ − sin 2θ cos 2θ + sin 2θ cos 2θ − sin 2θ 1 + tan 2θ = . 1 − tan 2θ Math 141 Lecture 6 Spring 2015 Trigonometric Integrals Integrating rational trigonometric integrals R The integral sec θdθ appears often in practice. A quicker solution will be shown later, but first we show the standard method. Example Set θ = 2 arctan t, cos θ = 1 − tan2 ( 2θ ) 1 1 − t2 , dθ = 2 dt. = 2 θ 2 1 + t 1 + t2 1 + tan 2 Z sec θdθ = ln |tan θ + sec θ| + C This is a perfectly good answer, however there’s a simplification: 2 sin 2θ cos 2θ + sin2 2θ + cos2 2θ sin θ + 1 tan θ + sec θ = = cos θ cos2 2θ − sin2 2θ 2 sin 2θ + cos 2θ sin 2θ + cos 2θ = = cos 2θ − sin 2θ cos 2θ + sin 2θ cos 2θ − sin 2θ 1 + tan 2θ = . 1 − tan 2θ Math 141 Lecture 6 Spring 2015 Trigonometric Integrals Ad hoc methods for trigonometric integrals Trigonometric Integrals - quick ad hoc techniques As we saw, every rational trigonometric expression can be integrated with the substitution θ = 2 arctan t. This integration technique results in rather long computations. Particular integral types may be computable with quicker ad hoc techniques. We illustrate such techniques on examples. Examples to which our ad hoc techniques apply arise from integrals needed outside of the subject of Calculus II, so these techniques are important. R dθ The trigonometric integral we saw, 2 sin θ−cos θ+5 , will not work with any of following ad-hoc techniques, so the general method is important as well. Math 141 Lecture 6 Spring 2015 Trigonometric Integrals Ad hoc methods for trigonometric integrals Example Z sin3 xdx = Z sin2 x sin xdx Z sin2 xd( − cos x) Z (−1) ?1 − cos2 x d(cos x) = Z = cos 2 x − 1 d( cos x) Z = u 2 − 1 du = u3 −u+C 3 1 = cos 3 x − cos x + C 3 Can we rewrite sin2 x via cos x? Set u = cos x = Math 141 Lecture 6 . Spring 2015 Trigonometric Integrals Ad hoc methods for trigonometric integrals Example Z 5 2 Z cos4 x sin2 x cos xdx Z Can we rewrite cos4 x sin2 xd( sin x) cos4 x via sin x? 2 cos2 x sin2 xd(sin x) 2 1 − sin 2 x sin 2 xd( sin x) Set u = sin x 2 1 − u 2 u 2 du 1 − 2u 2 + u 4 u 2 du u 2 − 2u 4 + u 6 du cos x sin xdx = = Z = Z = Z = Z = Z = u3 u5 u7 −2 + +C 3 5 7 3 5 sin x sin x sin 7 x = −2 + +C 3 5 7 = Math 141 Lecture 6 . Spring 2015 Trigonometric Integrals Z m Z n sin x cos xdx = m sin x cos n−1 Ad hoc methods for trigonometric integrals xd(sin x) n−1 2 d( sin x) sin m x 1 − sin 2 x Z n−1 2 du = um 1 − u2 Z = Z sinm x cosn xdx = Z When n − odd: cos xdx = d(sin x) Express cos x via sin x Set sin x = u When m − odd: sin xdx = d(− cos x) Express cos x cos n xd( cos x) via sin x sinm−1 x cosn xd( − cos x) m−1 2 1 − cos 2 x Z m−1 2 = − 1 − u2 u n du Set cos x = u 1−cos(2x) sin2 x = 2 If both m, n- even, use and substitute s = 2x cos2 x = cos(2x)+1 2 to lower trig powers. Repeat above considerations. =− Math 141 Z Lecture 6 Spring 2015 Trigonometric Integrals Ad hoc methods for trigonometric integrals Example Z 0 π 2 π 2 1 − cos(2x) sin xdx = dx 2 π 0 sin(2x) 2 x − = 4 0 2 π sin π sin 0 π = − = . − 0− 4 4 4 4 Z 2 express sin2 x via cos(2x) Example y y = p 1 − t2 t Set t = cos x, x ∈ 0, π2 ⇒ sin x ≥ 0. Then dt = d(cos x) = − sin xdx. Z t=1 p Z x=0 p 2 1 − t dt = − 1 − cos 2 x sin xdx π t=0 x= Z x= π 2p 2 = sin2 x sin xdx x=0 Z π 2 π = sin2 xdx = . 4 0 Math 141 Lecture 6 Spring 2015 Trigonometric Integrals Ad hoc methods for trigonometric integrals Example Z tan8 x sec4 xdx = Z tan8 x sec2 x sec2 xdx Z Can we rewrite tan8 x sec2 xd ( tan x) sec2 x via tan x? Z tan 8 x 1 + tan 2 x d( tan x) Set u = tan x = Z = u 8 1 + u 2 du Z = u 8 + u 10 du = u 9 u 11 + +C 9 11 9 tan x tan 11 x = + +C 9 11 = Math 141 Lecture 6 . Spring 2015 Trigonometric Integrals Ad hoc methods for trigonometric integrals Example Z 5 9 Z tan4 x sec8 x tan x sec xdx Z Can we rewrite tan4 x sec8 xd( sec x) tan4 x via sec x? 2 tan2 x sec8 xd(sec x) 2 sec 2 x − 1 sec 8 xd( sec x) Set u = sec x 2 1 − u 2 u 8 du 1 − 2u 2 + u 4 u 8 du u 8 − 2u 10 + u 12 du tan x sec xdx = = Z = Z = Z = Z = Z = u9 u 11 u 13 −2 + +C 9 11 13 9 11 sec x sec x sec 13 x = −2 + +C 9 11 13 = Math 141 Lecture 6 . Spring 2015 Trigonometric Integrals Ad hoc methods for trigonometric integrals Partial strategy for fast evaluation of Z m n Z tan x sec xdx = m tan x sec n−2 R tanm x secn xdx xd(tan x) n−2 2 tan x 1 + tan x d( tan x) = Z n−2 2 um 1 + u2 du = Z Z m 2 n − even, n ≥ 2 sec2 xdx = d(tan x) Express sec x via tan x Set u = tan x m − odd, n ≥ 1 tan x sec xdx tan x sec xdx = tan x sec xd(sec x) = d(sec x) Z m−1 Express tan x 2 = sec n−1 xd( sec x) sec 2 x − 1 via sec x Z m−1 2 u n du = u2 − 1 Set u = sec x m n Z m−1 n−1 Outside of the above cases we either use more tricks or resort to the general method x = 2 arctan t. Math 141 Lecture 6 Spring 2015 Trigonometric Integrals Ad hoc methods for trigonometric integrals Example Z Z tan xdx = = = Z sin x 1 dx = d( − cos x) cos x cos x Z du − = − ln |u| + C u − ln | cos x| + C = ln | sec x| + C Set u = cos x The following can be/was computed via x = 2 arctan t. Alternatively: Example Z Z sec xdx = = = = = Math 141 ( sec x + tan x) dx (sec x + tan x) Z sec2 x + sec x tan x dx sec x + tan x Z d( tan x + sec x) sec x + tan x Z du = ln |u| + C u ln | sec x + tan x| + C. sec x Lecture 6 Set u = sec x + tan x Spring 2015 Trigonometric Integrals Ad hoc methods for trigonometric integrals Example Z tan3 xdx Z = = = = = = = Math 141 tan x tan2 xdx Z 2 tan x sec x − 1 dx Z Z 2 tan x sec xdx − tan xdx Z tan xd( tan x) − ln | sec x| Z 1 udu + ln sec x u2 + ln | cos x| + C 2 tan 2 x + ln | cos x| + C 2 Lecture 6 Set u = tan x Spring 2015 Trigonometric Integrals Ad hoc methods for trigonometric integrals Example Z Z 3 sec xdx = sec x sec2 xdx Z = = = = = sec xd( tan x) Z sec x tan x − tan xd( sec x) Z sec x tan x − tan 2 x sec xdx Z sec x tan x − ( sec2 x − 1) sec xdx Z Z 3 sec x tan x − sec xdx + sec xdx Z sec3 xdx = sec x tan x + ln |sec x + tan x| + C Z sec3 xdx = 2 Math 141 Integrate by parts 1 (sec x tan x + ln |sec x + tan x|) + K . 2 Lecture 6 Spring 2015 Trigonometric Integrals Ad hoc methods for trigonometric integrals To evaluate integrals of the form R 1 sin mx cos nxdx R 2 sin mx sin nxdx R 3 cos mx cos nxdx use the corresponding identity: 1 sin A cos B = 12 [sin(A − B) + sin(A + B)] 2 sin A sin B = 12 [cos(A − B) − cos(A + B)] 3 cos A cos B = 12 [cos(A − B) + cos(A + B)] Math 141 Lecture 6 Spring 2015 Trigonometric Integrals Ad hoc methods for trigonometric integrals Example Z Z sin 4x cos 5xdx = = = = Math 141 1 [sin(4x − 5x) + sin(4x + 5x)]dx 2 Z 1 (sin(−x) + sin(9x))dx 2 Z 1 (− sin x + sin(9x))dx 2 1 1 (cos x − cos(9x)) + C 2 9 Lecture 6 Spring 2015